Definition 2.2.1.
The set of complex numbers is denoted by \(\mathbb{C}\) and is described by \(\mathbb{C} = \left\{ \ \frac{z}{z} = x + \text{iy},\ x,\ y \in \Re\ \text{and}\ i^{2} = - 1\ \right\}\text{.}\)
Section 2.2 The set of complex numbers
The positive integers (natural numbers) were invented to count things. The negative integers were introduced to count money when we owed more than we had. The rational numbers were invented for measuring quantities. Since quantities like voltage, length and time can be measured using fractions, they can be measured using the rational numbers.
The real numbers were invented for wholly mathematical reasons: it was found that there were lengths such as the diagonal of the unit square which, in principle, couldnβt be measured by the rational numbers, instead they can be measured using real numbers.
The complex numbers were invented for purely mathematical reasons, just like the real numbers and were intended to make things neat and tidy in solving equations. They were regarded with deep suspicion by the more conservative folk for a century. Complex numbers are points in the plane, together with a rule telling you how to multiply them. They are two-dimensional, whereas the real numbers are one dimensional.
Equations of the form \(x^{2} + 1 = 0\)has no solution on the set of real numbers. Therefore, the set of complex numbers permits us to solve such equations.
Note: If \(x = 0\text{,}\) the number is called purely imaginary and if \(y = 0\text{,}\) the number is called purely real.
Complex numbers can be defined as an order pair \((x, y)\) of real numbers that can be interpreted as points in the complex plane (z- plane) with coordinates \(x\) and \(y\text{.}\)
Example 2.2.2.
Find the real \(\&\) imaginary part of the following complex numbers :
\(\ \text{(a)}\ \text{z} = 3 + 7i\)
\(\ \text{(b)}\ \text{z} = 1 - i\)
Subsection 2.2.1 Plotting complex numbers
Any complex number \(z = x + iy \) can be drawn in the complex plane as below :
Example 2.2.3.
Draw the complex number \(z = 2+3i\)
Solution.
\begin{equation*}
\textbf{Equality of Complex numbers}
\end{equation*}
Two complex numbers \(z_1 = a + ib\) and \(z_2 = c + id\) are equal iff \(a = c \) & \(b = d \text{.}\)
Example 2.2.4.
If \(z_1 = 2 + ix\) and \(z_2 = y + 6i\) are equal, then find the value of \(x\) & \(y\text{.}\)
Checkpoint 2.2.5.
Checkpoint 2.2.6.
Subsection 2.2.2 Operations on Complex numbers
Example 2.46\(\ \text{Let}\ z_{1} = a + ib\ \text{and}\ z_{2} = c + id,\ \text{be}\ \text{any}\ \text{two}\ \text{complex}\ \text{numbers},\ \text{then}\)
\(\ i)\ z_{1} + z_{2} = (a + c) + i (b + d )\)
\(\ ii)\ z_{1} - z_{2} = (a - c) + i (b - d )\)
\(\ iii)\ z_{1}\text{.}\ z_{2} = (a + ib) \text{.} (c + id) = a(c + id ) + ib(c + id ) = ac + iad + ibc - bd = (ac - bd ) + i (ad + bc)\)
\(\ iv)\ \frac{z_{1}}{z_{2}} = \frac{(a + ib)}{(c + id )} \text{,} \ z_2 \neq 0\text{.}\)
Example 2.2.7.
If\(\ z_{1} = 2 + 3i\) and \(\ z_{2} = 4 + i\text{,}\) then find \(a)\ z_{1} + z_{2} \ b)\ \ z_{1} - z_{2}\ \ c)\ z_{1}\text{.}\ z_{2}\ \ d)\ \frac{z_{1}}{z_{2}}\)
Checkpoint 2.2.8.
Checkpoint 2.2.9.
Subsection 2.2.3 Conjugate of a complex number
Definition 2.2.10.
The conjugate of a complex number z = x+iy is denoted by \(\ \overline{z}\) and is defined as \(\ \overline{z}\) = x-iy. It can be represented by the point (x, -y) which is the reflection of the point (x, y) about the x-axis.
Example 2.2.11.
Find the conjugate of the complex number \(z = 2+9i\text{.}\)
\begin{equation*}
\textbf{Properties of Conjugate}
\end{equation*}
\(a. \overline{\overline{z}} = z\)
\(b. z + \overline{z} = 2x =2Re{(z)} = 2 \left (\frac{z + \overline{z}}{2} \right )\)
\(c. z - \overline{z} = 2iy = 2Im{(z)}\)
\(d. \overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}\)
\(e. \overline{z_1 - z_2} = \overline{z_1} - \overline{z_2}\)
\(f. \overline{z_1 \cdot z_2} = \overline{z_1} \cdot \overline{z_2}\)
\(g. \overline{({\frac{z_1}{z_2}})} = \frac{\overline{z_1}}{\overline{z_2}}\)
proof : a) let \(z = x + iy\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow \overline{z} = x - iy\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow \overline{\overline{z}} = x + iy = z\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \ \overline{\overline{z}} = z\text{.}\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ d) \ \text{Let} \ z_{1} = x_{1} + iy_{1} \) & \(z_{2} = x_{2} + iy_{2}\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow \overline{z_{1}} = x_{1} - iy_{1} \ \ \ \) & \(\ \ z_{2} = x_{2} - iy_{2}\)
Now \(\overline{z_1 + z_2} = \overline{(x_{1} + iy_{1}) + (x_{2} + iy_{2})} = \overline{(x_{1} + x_{2}) + i(y_{1} + y_{2})} = (x_{1} + x_{2}) + i(y_{1} + y_{2})\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = x_{1} - iy_{1} + x_{2} - iy_{2}\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \overline{z_1} + \overline{z_2}\)
The others are left for the reader.
Checkpoint 2.2.12.
Checkpoint 2.2.13.
Subsection 2.2.4 Modulus (Norm) of a complex number
Definition 2.2.14.
The modulus of a complex number z = x+iy is a non-negative real number denoted by \(\left |z \right |\) and is defined as \(\left |z \right |\) = \(\sqrt{x^{2} + y^{2}}\text{.}\) Geometrically, the number \(\left |z \right |\) represents the distnce between the point (x, y) and the origin.
Example 2.2.15.
Find the modulus of the complex number \(z = 3 - 4i\text{.}\)
\begin{equation*}
\textbf{Properties of modulus}
\end{equation*}
b. \(\left | z^{2} \right | = z \cdot \overline{z}\)
c. \(\left |z_{1} \cdot z_{2} \right | = \left |z_{1} \right | \cdot \left |z_{2} \right |\)
d. \(\left | \frac{z_{1}}{z_{2}} \right | = \frac{\left |z_{1} \right |}{\left |z_{2} \right |}\ \ \ \ \ \ \ \ \) (if \(z_{2} \neq 0\))
e. \(\left |z_{1} + z_{2} \right | \leq \left |z_{1} \right | + \left |z_{2} \right | \ldots \ldots\ldots\ldots\ldots\) triangle inequality
f. \(\left |z_{1} - z_{2} \right | \geq \left |z_{1} \right | - \left |z_{2} \right |\)
proof \((a) \ \ \text{let} \ \ \ z = x + iy \ \ \ \ \text{from} \ \ \ \text{which} \ \ \ \ \ \overline{z} = x - iy\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow \left |z \right | = \sqrt{x^{2} + y^{2}} \ \ \ \ \ \ \ \ \ \ \ \Rightarrow \left |\overline{z} \right | = \sqrt{x^{2} + y^{2}}\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \left |z \right | = \left |\overline{z} \right |\)
proof \((b) \ \ \text{let} \ \ \ z = x + iy \ \ \ \ \text{from} \ \ \ \text{which} \ \ \ \ \ \overline{z} = x - iy\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow \left |z \right |^{2} = x^{2} + y^{2}\)
Now \(\ \ \ \ z\cdot\overline{z} = (x - iy) \cdot (x - iy) = x^{2} + y^{2}\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \left |z \right |^{2} = z \cdot \overline{z}\)
proof \((c) \ \ \ \left |z_{1} \cdot z_{2} \right |^{2} = (z_{1} \cdot z_{2})(\overline{{z_{1}} \cdot z_{2}}) = (z_{1} \cdot z_{2})(\overline{z_{1}} \cdot \overline{z_{2}}) = z_{1} \cdot \overline{z_{1}} \cdot z_{2} \cdot \overline{z_{2}}\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \left |z_{1} \right |^{2} \cdot \left |z_{2} \right |^{2}\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (\left |z_{1} \right | \cdot \left |z_{2} \right |)^{2} \cdots\cdots\cdots\cdots\cdots from \ \ (b)\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow \left |z_{1} \cdot z_{2} \right | = \left |z_{1} \right | \cdot \left |z_{2} \right |\)
proof \((d) \ \ \ \left | \frac{z_{1}}{z_{2}} \right | = \sqrt{\frac{z_{1} \cdot \overline{z_{1}}}{z_{2} \cdot \overline{z_{2}}}} = \sqrt{\frac{\left |z_{1} \right |^{2}}{\left |z_{2} \right |^{2}}} = \frac{\left |z_{1} \right |}{\left |z_{2} \right |}\) (if \(z_{2} \neq 0\))
The others are left for the reader.
Checkpoint 2.2.16.
Checkpoint 2.2.17.
Subsection 2.2.5 Additive and multiplicative inverses
Let z = x+iy be a complex number, then
-
its additive inverse denoted by \((- z )\) is given by : \(-z = - ( x + iy) = - x - iy\text{.}\)
-
its multiplicative inverse denoted by \(z ^{-1}\) is given by : \(z^{-1} = \frac{1}{x + iy} = \frac{x}{x^{2} + y^{2}} - \frac{iy}{x^{2} + y^{2}}\text{.}\)
Example 2.2.18.
Find the additive and multiplicative inverses of the complex number \(z = 3 + 4i\text{.}\)
Exercises Exercises
1.
Verify that
-
\(\displaystyle (\sqrt{2} - i) - i (1 - \sqrt{2i}) = -2i\)
-
\(\displaystyle (2, -3)(-2, 1) = (-1, 8)\)
-
\(\displaystyle (3, 1)(3, -1)(\frac{1}{5}, \frac{1}{10}) = (2, 1)\)
-
\(\displaystyle (2 + 3i)^2 - (3i - 6) = 1+9i\)
2.
Show that
-
\(\displaystyle Re(iz) = -Im(z)\)
-
\(\displaystyle Im(iz) = Re(z)\)
-
\(\displaystyle (z + 1)^{2} = z^{2} + 2z + 1\)
3.
Do the following operations and simplify your answer.
-
\(\displaystyle \frac{1 + 2i}{3 - 4i} + \frac{2 - i}{5i}\)
-
\(\displaystyle \frac{5i}{(1 - i)(2 - i)(3 - i)}\)
-
\(\displaystyle (1 - i)^{3}\)
Checkpoint 2.2.19.
Subsection 2.2.6 Argument (Amplitude) of a complex number
Definition 2.2.20.
Argument of a complex number \(z = x+iy\) is the angle formed by the complex number \(z = x+iy\) with the positive x-axis. The argument of a complex number \(z = x+iy\) is deonted by \(\textbf{argz}\) and is given by arg(z) = \(\text{tan}^{- 1}(\frac{y}{x})\text{.}\)
The particular argument of \(z\) that lies in the range \(- \pi < \theta \leq \pi\) is called the principal argument of \(z\) and is dented by Argz.
Notes : \(i) Argz \ \ \in \left ( -\pi , \pi \right ]\text{.}\)
\(\ \ \ \ \ \ \ \ \ \ \ \ ii) \text{If} \ 0 \leq \ \text{Argz} \ \leq \pi,\) move counter clock wise direction, if not move the other direction.
Example 2.2.21.
Find the argument of the following complex numbers:
-
\(\displaystyle z = 1 + i\)
-
\(\displaystyle z = - 2 + 2 \sqrt{3i}\)
-
\(\displaystyle z = - \sqrt{3} - i\)
Solution.
\(a) \ \ z = 1 + i\)
\(\text{Arg}z = \text{tan}^{- 1}\left(\frac{1}{1}\right) = \text{tan}^{- 1}\left(1 \right) = \frac{\pi}{4}\)
\(b) \ \ z = - 2 + 2 \sqrt{3}i\)
\(\text{Arg}z = \text{tan}^{- 1}\left(\frac{2 \sqrt{3}}{-2}\right) = \text{tan}^{- 1}(-\sqrt{3}) = \frac{2 \pi}{3}\)
\(c) \ \ z = - \sqrt{3} - i\)
\(\text{Arg}z = \text{tan}^{- 1}\left(\frac{-1}{-\sqrt{3}}\right) = \text{tan}^{- 1}\left(\frac{1}{\sqrt{3}}\right) = \frac{-5 \pi}{6}\)
\begin{equation*}
\textbf{Properties of Arguments}
\end{equation*}
\(i) \ \ \text{Arg}(z_{1} \cdot z_{2}) = \text{Arg}z_{1} \ + \ \text{Arg}z_{2}\)
\(ii) \ \ \text{Arg}\left(\frac{z_{1}}{z_{2}}\right) = \text{Arg}z_{1} \ - \ \text{arg}z_{2}\)
Example 2.2.22.
Find the principal argument of \(\ a)\ (1 + i)\ ( - 1 - i)\ \ \ \ b)\ (\frac{- 2 + 2i}{1 - i})\)
Solution.
\(a) \ \ \text{Arg} \ (1 + i)(-1 - i) = \text{Arg} \ (1 + i) + \text{Arg} \ (-1 - i) = \frac{\pi}{4} + (\frac{-3 \pi}{4}) = \underline{\underline{\frac{-\pi}{2}}} \)
\(b) \ \ \text{Arg} \left(\frac{- 2 + 2i}{1 - i}\right) = \text{Arg}(-2 + 2i) - \text{Arg}(1 - i) = (\frac{3 \pi}{4}) - (-\frac{\pi}{4}) = \frac{4 \pi}{4} = \underline{\underline{\pi}}\)
Checkpoint 2.2.23.
Checkpoint 2.2.24.
Subsection 2.2.7 Polar form of a complex number
Definition 2.2.25.
Let \(r\) and \(\theta\) be polar coordinates of the point (x, y) of the complex number \(z = x+iy\text{.}\) Since x = \(r\text{cos}\theta\)and y = \(r\text{sin}\theta\text{,}\) then the complex number can be written as : \(\underline{\underline{\ z = r(\text{cos}\theta + i\text{sin}\theta)}}\) which is called polar form, where \(r\) is modulus of \(z\) and \(\ \theta\ \)is principal argument of \(z\text{.}\)
Example 2.2.26.
Express the following complex numbers in polar form:
\(a)\ z = 1 + i \)
\(\text{solution}:\ r = \sqrt{2}\ \text{and}\ \theta = \text{tan}^{- 1}(1) = \frac{\pi}{4} \text{.}\)
\(\ \text{Thus},\ z = \sqrt{2}(\text{cos} \frac{\pi}{4} + i\text{sin} \frac{\pi}{4})\text{.}\)
\(b)\ z = 3 - 3i\ \text{solution}:\ r = \sqrt{18}\ \text{and}\ \theta = \text{tan}^{- 1}( - 1) = - \frac{\pi}{4}\text{.}\)
\(\text{solution}: \ r = \sqrt{\text{18}}\ \text{and}\ \theta = \text{tan}^{- 1}( - 1) = - \pi 4\text{.}\ \\\)
\(\text{Thus},\ z = \sqrt{\text{18}}(\text{cos} - \frac{\pi}{4} + i\text{sin} - \frac{\pi}{4}) = \sqrt{\text{18}}(\text{cos}\frac{\pi}{4} - i\text{sin}\frac{\pi}{4})\text{.} \\\)
\begin{equation*}
\textbf{Multiplication and division in polar forms}
\end{equation*}
\(\text{If} \ z_{1}\ = r_{1}(\text{cos}\theta_{1} + i\text{sin}\theta_{1}) \ \ \ \text{and} \ \ \ \ z_{2} = r_{2}(\text{cos}\theta_{2} + i\text{sin}\theta_{2}), \text{then}\)
\(a) \ \ z_{1} \cdot z_{2} = r_{1} \cdot r_{2}(\text{cos}(\theta_{1} + \theta_{2}) + i\text{sin}(\theta_{1} + \theta_{2}))\)
\(b) \ \ \frac{z_{1}}{z_{2}} = \frac{r_{1}} {r_{2}}(\text{cos}(\theta_{1} - \theta_{2}) + i\text{sin}(\theta_{1} - \theta_{2}))\text{.}\)
\(\textbf{Proof:}\)
\(\text{a)} \ z_{1} \cdot z_{2} = r_{1}(\text{cos}\theta_{1} + i\text{sin}\theta_{1}) \cdot r_{2}(\text{cos}\theta_{2} + i\text{sin}\theta_{2})\)
\(\ \ \ \ \ \ \ \ \ \ \ = r_{1} \cdot r_{2} \left [ \text{cos}\theta_{1}(\text{cos}\theta_{2} + i\text{sin}\theta_{2}) + i\text{sin}\theta_{1}(\text{cos}\theta_{2} + i\text{sin}\theta_{2}) \right ]\)
\(\ \ \ \ \ \ \ \ \ \ \ = r_{1} \cdot r_{2} \left [ \text{cos}\theta_{1}\text{cos}\theta_{2} + i\text{cos}\theta_{1}\text{sin}\theta_{2} + i\text{sin}\theta_{1}\text{cos}\theta_{2} - \text{sin}\theta_{1}\text{sin}\theta_{2} \right ]\)
\(\ \ \ \ \ \ \ \ \ \ \ = r_{1} \cdot r_{2} \left [ \text{cos}\theta_{1}\text{cos}\theta_{2} - \text{sin}\theta_{1}\text{sin}\theta_{2} + i(\text{cos}\theta_{1}\text{sin}\theta_{2} + \text{sin}\theta_{1}\text{cos}\theta_{2}) \right ]\)
\(\ \ \ \ \ \ \ \ \ \ \ = \underline{\underline{ r_{1} \cdot r_{2} \left [ \text{cos}(\theta_{1} + \theta_{2}) + i \ \text{sin}(\theta_{1} + \theta_{2})\right ]}}\)
\(\text{Thus,} \ z_{1} \cdot z_{2} = r_{1} \cdot r_{2}(\text{cos}(\theta_{1} + \theta_{2}) + i \ \text{sin}(\theta_{1} + \theta_{2}))\)
\(\text{b)} \ \frac{z_{1}}{z_{2}} = \frac{r_{1} (\text{cos}\theta_{1} + i\text{sin}\theta_{1})}{ r_{2} (\text{cos}\theta_{2} + i\text{sin}\theta_{2})} = \frac{r_{1}}{r_{2}} \left ( \frac{\text{cos}\theta_{1} + i\text{sin}\theta_{1}}{\text{cos}\theta_{2} + i\text{sin}\theta_{2}} \right )\)
\(\ \ \ \ \ \ \ \ = \frac{r_{1}}{r_{2}} \left ( \frac{\text{cos}\theta_{1} + i\text{sin}\theta_{1}}{\text{cos}\theta_{2} + i\text{sin}\theta_{2}} \right ) \cdot \frac{\text{cos}\theta_{2} - i\text{sin}\theta_{2}}{\text{cos}\theta_{2} - i\text{sin}\theta_{2}}\)
\(\ \ \ \ \ \ \ \ = \frac{r_{1}}{r_{2}} \left ( \frac{ \text{cos}\theta_{1}\text{cos}\theta_{2} - i\text{cos}\theta_{1}\text{sin}\theta_{2} + i\text{sin}\theta_{1}\text{cos}\theta_{2} + \text{sin}\theta_{1}\text{sin}\theta_{2} }{\text{cos}^{2}\theta_{2} + \text{sin}^{2}\theta_{2}}\right )\)
\(\ \ \ \ \ \ \ \ = \frac{r_{1}}{r_{2}} \left ( \frac{\text{cos}\theta_{1}\text{cos}\theta_{2} + \text{sin}\theta_{1}\text{sin}\theta_{2} + i(\text{sin}\theta_{1}\text{cos}\theta_{2} - \text{cos}\theta_{1}\text{sin}\theta_{2})}{1} \right )\)
\(\ \ \ \ \ \ \ \ = \underline{\underline{ \frac{r_{1}}{r_{2}} \left [ \text{cos}(\theta_{1} - \theta_{2}) + i\text{sin}(\theta_{1} - \theta_{2})\right ]}}\)
\(\text{Thus,} \ \frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}}(\text{cos}(\theta_{1} - \theta_{2}) + i\text{sin}(\theta_{1} - \theta_{2}))\)
Example 2.2.27.
If \(z_{1} = 6(\text{cos}\frac{\pi}{2} + i\text{sin} \frac{\pi}{2})\) and \(z_{2} = 2(\text{cos}\frac{\pi}{3} + i\text{sin}\frac{\pi}{3})\text{,}\) then find:
\(a) \ z_{1} \cdot z_{2} \ \ \ \ \ b) \ \frac{z_{1}}{z_{2}}\)
Solution.
\(\text{a) } z_{1} \cdot z_{2} = 6 \cdot 2 \left [ \text{cos}\left(\frac{\pi}{2} + \frac{\pi}{3}\right) + i\text{sin}\left(\frac{\pi}{2} + \frac{\pi}{3}\right) \right ]\)
\(\ \ \ \ \ \ \ \ = \underline{\underline{ 12\left [ \text{cos}\left(\frac{5\pi}{6}\right) + i\text{sin}\left(\frac{5\pi}{6}\right) \right ]}}\)
\(\text{b) } \frac{z_{1}}{z_{2}} = \frac{6}{2} \left [ \text{cos}\left(\frac{\pi}{2} - \frac{\pi}{3}\right) + i\text{sin}\left(\frac{\pi}{2} - \frac{\pi}{3}\right) \right ]\)
\(\ \ \ \ \ \ \ \ = \underline{\underline{ 3\left [ \text{cos}\left(\frac{\pi}{6}\right) + i\text{sincos}\left(\frac{\pi}{6}\right) \right ]}}\)
\(\text{Thus, } z_{1} \cdot z_{2} = 12\left [ \text{cos}\left(\frac{5\pi}{6}\right) + i\text{sin}\left(\frac{5\pi}{6}\right) \right ] \text{ and } \frac{z_{1}}{z_{2}} = 3\left [ \text{cos}\left(\frac{\pi}{6}\right) + i\text{sin}\left(\frac{\pi}{6}\right) \right ]\)
Definition 2.2.28.
\begin{equation*}
\textbf{Argument of a product}
\end{equation*}
The argument of the product of two complex numbers is the sum of their arguments.
\(\textbf{Proof:}\)
\(\text{Let } z_{1} = r_{1}(\text{cos}\theta_{1} + i\text{sin}\theta_{1}) \ \ \ \text{and} \ \ \ z_{2} = r_{2}(\text{cos}\theta_{2} + i\text{sin}\theta_{2})\)
\(\text{Now, } z_{1} \cdot z_{2} = r_{1}(\text{cos}\theta_{1} + i\text{sin}\theta_{1}) \cdot r_{2}(\text{cos}\theta_{2} + i\text{sin}\theta_{2})\)
\(\ \ \ \ \ \ \ \ = r_{1} \cdot r_{2} \left [ \text{cos}\theta_{1}(\text{cos}\theta_{2} + i\text{sin}\theta_{2}) + i\text{sin}\theta_{1}(\text{cos}\theta_{2} + i\text{sin}\theta_{2}) \right ]\)
\(\ \ \ \ \ \ \ \ = r_{1} \cdot r_{2} \left [ \text{cos}\theta_{1}\text{cos}\theta_{2} + i\text{cos}\theta_{1}\text{sin}\theta_{2} + i\text{sin}\theta_{1}\text{cos}\theta_{2} - \text{sin}\theta_{1}\text{sin}\theta_{2} \right ]\)
\(\ \ \ \ \ \ \ \ = r_{1} \cdot r_{2} \left [ \text{cos}\theta_{1}\text{cos}\theta_{2} - \text{sin}\theta_{1}\text{sin}\theta_{2} + i(\text{cos}\theta_{1}\text{sin}\theta_{2} + \text{sin}\theta_{1}\text{cos}\theta_{2}) \right ]\)
\(\ \ \ \ \ \ \ \ = \underline{\underline{ r_{1} \cdot r_{2} \left [ \text{cos}(\theta_{1} + \theta_{2}) + i\text{sin}(\theta_{1} + \theta_{2})\right ]}}\)
\(\Rightarrow \text{arg} (z_{1} \cdot z_{2}) = \text{tan}^{-1} \left ( \frac{\text{sin}(\theta_{1} + \theta_{2})}{\text{cos}(\theta_{1} + \theta_{2})} \right )\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \text{tan}^{-1} \left ( \text{tan}(\theta_{1} + \theta_{2}) \right )\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \underline{\underline{ \theta_{1} + \theta_{2}}}\)
Definition 2.2.29.
\begin{equation*}
\textbf{Argument of the quotient}
\end{equation*}
The argument of the quotient of two complex numbers is the difference of their arguments.
\(\textbf{Proof:}\)
\(\text{arg} \left ( \frac{z_{1}}{z_{2}} \right ) = \text{arg}(z_{1} \cdot {z_{2}}^{-1}) = \text{arg}(z_{1}) + (\text{arg} z_{2}^{-1}) = \text{arg}(z_{1}) + - (\text{arg} z_{2}) = \text{arg}(z_{1}) - \text{arg}(z_{2})\)
Example 2.2.30.
\(\text{arg} \left ( \frac{-4}{1 + \sqrt{3}\ i} \right ) = \text{arg}(-4) - \text{arg}(1 + \sqrt{3}\ i) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\)
\begin{equation*}
\textbf{De Moivre's Formula}
\end{equation*}
Recall the product :\(\ z_{1} \cdot z_{2} = r_{1} \cdot r_{2} \left [ \text{cos}(\theta_{1} + \theta_{2}) + i\text{sin}(\theta_{1} + \theta_{2})\right ]\text{.}\)
Similarly, we get \(z_{1} \cdot z_{2} \cdots. z_{n} = r_{1} \cdot r_{2} \cdots. r_{n} \left [ \text{cos}(\theta_{1} + \theta_{2} + \cdots + \theta_{n}) + i\text{sin}(\theta_{1} + \theta_{2} + \cdots + \theta_{n})\right ]\text{.}\)
Now we can generalize that\(z^{n} = r^{n} \left [ \text{cos}(\theta + \theta + \cdots + \theta) + i\text{sin}(\theta + \theta + \cdots + \theta)\right ]\text{.}\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \underline{\underline{ r^{n} \left [ \text{cos}(n\theta) + i\text{sin}(n\theta)\right ]}}\) which is called De Moivreβs formula.
Example 2.2.31.
Express \((2 + 2i)^{100}\) in polar form.
Solution.
\(\text{Let } z = 2 + 2i. Then, r = \sqrt{8} , \ \ \theta = \frac{\pi}{4} \) and hence
\((2 + 2i)^{100} = z^{100}\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \sqrt{8}^{100} \left [ \text{cos} 100 (\frac{\pi}{4}) + i\text{sin} 100 (\frac{\pi}{4})\right ]\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \underline{\underline{ 8^{50} \left [ \text{cos}25\pi + i\text{sin}25\pi \right ]}}\)
Example 2.2.32.
Express \((\sqrt{3} + i)^{60}\) in polar form.
Solution.
\(\text{Let } z = \sqrt{3} + i. Then, r = 2 , \ \ \theta = \frac{\pi}{6} \text{.}\)
\(\therefore (\sqrt{3} + i)^{60} = z^{60}\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2^{60} \left [ \text{cos} 60 (\frac{\pi}{6}) + i\text{sin} 60 (\frac{\pi}{6})\right ]\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \underline{\underline{ 2^{60} \left [ \text{cos}10\pi + i\text{sin}10\pi \right ]}}\)
Definition 2.2.33.
\begin{equation*}
\textbf{Euler's formula}
\end{equation*}
The complex number \(z = r(\text{cos}\theta + i\text{sin}\theta)\) can be written in exponential form as: \(z = re^{i\theta}\) which is called Eulerβs formula.
\(\text{Note:} \ \ \ \ z^{n} = r^{n} \left ( \text{cos}n \theta + i \text{sin}n\theta \right ) = r^{n} e^{{i}(n\theta)}\)
Example 2.2.34.
\(\text{Express the complex number } z = 1 + i \text{ using Euler's formula.}\)
Example 2.2.35.
\(\text{Express the complex number } z = 1 + \sqrt{3}i \text{ using Euler's formula.}\)
Example 2.2.36.
\(\text{Express the complex number } z = (\sqrt{3} + i)^{7} \text{ using Euler's formula.}\)
Checkpoint 2.2.37.
Checkpoint 2.2.38.
Checkpoint 2.2.39.
Checkpoint 2.2.40.
Checkpoint 2.2.41.
Checkpoint 2.2.42.
Subsection 2.2.8 Extraction of roots
Suppose \(z_{0} = r_0\text{e}^{\text{iΞΈ}_{0}}\ \) is the \(n^{th}\) root of a non - zero complex number \(z = re^{i \theta}\text{,}\) where \(n \geq 2\text{.}\)
Then, \(z_{0}^{n} = z\text{,}\) which implies that \(r_0^{n}\text{e}^{\text{i nΞΈ}_{0}} = re^{i \theta}\)
\(\ \ \ \ \ \ \ \Rightarrow r_0^{n} = r\) & \(\ n\theta_0 \ = \ \ \theta + 2k\pi, \ \ \ \ k = 0, 1, 2, \cdots (n - 1)\text{.}\)
\(\ \ \ \ \ \ \ \Rightarrow r_0 = (r)^{\frac{1}{n}} \ \) & \(\ \theta_0 \ = \frac{\theta}{n} + \frac{2k\pi}{n}\)
\(\therefore, \ \ \ z_{0} = \underline {\underline { (r)^{\frac{1}{n}} \left ( \text{e}^{\text{i}\left(\frac{\theta}{n} + \frac{2k\pi}{n}\right)} \right)}} \ ,\) which is the \(n^{th}\) root of \(z\text{,}\) where \(n = 2, 3, \cdots\) and \(k = 0, 1, 2, \cdots (n - 1)\) or we can denote it by \(C_{k} \ \ \ \text{as :} \ \ C_{k} = \underline {\underline { (r)^{\frac{1}{n}} \left( \text{e}^{\text{i}\left(\frac{\theta}{n} + \frac{2k\pi}{n}\right)} \right)}} \ ,\) where \(k = 0, 1, 2, \cdots (n - 1)\text{.}\)
Example 2.2.43.
Find the square roots of the complex number \(z = 1 + \sqrt{3}i\text{.}\)
Solution.
\(z = 1 + \sqrt{3}i\)
Here \(\ \ r = 2, \theta = \frac{\pi}{3}\)
Hence \(C_{k} = (r)^{\frac{1}{n}} \left( \text{e}^{\text{i}\left(\frac{\theta}{n} + \frac{2k\pi}{n}\right)} \right), n = 2\ , \ k = 0, 1\text{.}\)
\(\ \ \ \ \ \ \ \Rightarrow C_{k} = (2)^{\frac{1}{2}} \left ( \text{e}^{\text{i}\left(\frac{\frac{\pi}{3}}{2} + \frac{2k\pi}{2} \right )} \right )\)
\(\ \ \ \ \ \ \ \Rightarrow C_{k} = \sqrt{2} \left( \text{e}^{\text{i}\left(\frac{\pi}{6} + k\pi\right)} \right)\)
\(i.) \ \ \text{If} \ \ k = 0\text{,}\) then \(C_{0} = \sqrt{2} \left( \text{e}^{\text{i}\frac{\pi}{6}} \right) = \sqrt{2} \left( \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} \right) = \sqrt{2} \left ( \frac{\sqrt{3}}{2} + \frac{i}{2} \right ) = \frac{\sqrt{2}}{2} \left ( \sqrt{3} + i \right ) = \underline {\underline {\frac{\sqrt{6}}{2} + \frac{\sqrt{3i}}{2}}}\)
\(ii.) \ \ \text{If} \ \ k = 1\text{,}\) then \(C_{1} = \sqrt{2} \left( \text{e}^{\text{i}\left(\frac{\pi}{6} + \pi\right)} \right) = \sqrt{2} \left ( \text{cos}\frac{7\pi}{6} + i\text{sin}\frac{7\pi}{6} \right ) = \sqrt{2} \left( \frac{-\sqrt{3}}{2} - \frac{i}{2} \right) = \frac{-\sqrt{2}}{2} \left( \sqrt{3} + i \right) = \underline {\underline {\frac{-\sqrt{6}}{2} - \frac{\sqrt{3i}}{2}}}\)
\(\therefore \ \ \ \text{The square roots of} \ \ \ 1 + \sqrt{3}i \ \text{are} \ \ \ C_{0} = \frac{\sqrt{6}}{2} + \frac{\sqrt{3}i}{2} \ \ \) & \(\ \ C_{1} = \frac{- \sqrt{6}}{2} - \frac{\sqrt{3i}}{2}\text{.}\)
Example 2.2.44.
Find the cube roots of the complex number \(z = 8i\text{.}\)
Solution.
Hence, \(C_{k} = (r)^{\frac{1}{n}} \left( \text{e}^{\text{i}\left(\frac{\theta}{n} + \frac{2k\pi}{n}\right)} \right)\)
\(\ \ \ \ \ \ \ \Rightarrow C_{k} = (8)^{\frac{1}{3}} \left( \text{e}^{\text{i}\left(\frac{\frac{\pi}{2}}{3} + \frac{2k\pi}{3}\right)} \right)\)
\(\ \ \ \ \ \ \ \Rightarrow C_{k} = 2 \left( \text{e}^{\text{i}\left(\frac{\pi}{6} + \frac{2k\pi}{3}\right)} \right)\)
\(i.) \ \ \text{If} \ \ k = 0\text{,}\) then \(C_{0} = 2 \left( \text{e}^{\text{i}\frac{\pi}{6}} \right) = 2 \left( \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} \right) = 2 \left( \frac{\sqrt{3}}{2} + \frac{i}{2} \right) = \underline {\underline {\sqrt{3} + i}}\ \)
\(ii.) \ \ \text{If} \ \ k = 1\text{,}\) then \(C_{1} = 2 \left( \text{e}^{\text{i}\left(\frac{\pi}{6} + \frac{2\pi}{3}\right)} \right) = 2 \left( \cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6} \right) = 2 \left( -\frac{\sqrt{3}}{2} + \frac{i}{2} \right) = \underline {\underline {- \sqrt{3} + i}}\ \)
\(iii.) \ \ \text{If} \ \ k = 2\text{,}\) then \(C_{2} = 2 \left( \text{e}^{\text{i}\left(\frac{\pi}{6} + \frac{4\pi}{3}\right)} \right) = 2 \left( \cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2} \right) = 2(0 + - i) = \underline {\underline {- 2i}}\ \)
\(\therefore \ \ \ \text{The cube roots of} \ \ 8i \ \text{are} \ \ C_{0} = \sqrt{3} + i, C_{1} = -\sqrt{3} + i\) & \(C_{2} = -2i\text{.}\)
Exercises Exercises
1.
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Find the square roots of the complex numbers:\(\ \text{(a)}\ \text{z} = \frac{3i}{-1 - i}\)\(\ \text{(b)}\ \text{z} = \left ( \sqrt{3} - i \right )\)
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Show that \(\ \text{a)} \ \ \left | \text{e}^{i\theta} \right | = 1 \ \ \ \ \ \ \ \ \ \ \ \text{b)} \ \ \ \overline {\text{e}^{i\theta}} = \text{e}^{-i\theta} \)
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Using mathematical induction, show that \(\ \text{e}^{i\theta_{1}} \cdot \text{e}^{i\theta_{2}}. \cdots . \text{e}^{i\theta_{n}} = \text{e}^{i(\theta_{1} + \theta_{2} + \cdots + \theta_{n})} \) \(n = 2, 3, \cdots\)
