Learning Objectives:

1. Recognize the need and origin of imaginary numbers.
2. Identify real and imaginary part of a complex number.

Concept of imaginary numbers

\[j = \sqrt{-1}\]

Powers of imaginary number

\[\begin{aligned} j^2 &= -1\\ j^3 &= j^2 \times j = -j\\ j^4 &= j^2 \times j^2 = -1 \times -1 = 1\end{aligned}\]

After this , the same cycle repeats. \[\begin{aligned} j^5 &= j^4 \times j = j\\ j^6 &= j^4 \times j^2 = -1 \end{aligned}\]

and so on..

Solved Example: 9170-01

The argument of the complex number $\dfrac{1+i}{1-i}$, where $i= \sqrt{-1}$, is:

A. $-\pi$

B. $\dfrac{-\pi}{2}$

C. $\dfrac{\pi}{2}$

D. $\pi$

We can rationalize the given expression to bring it in the standard a + ib format.\\ \begin{align*} \dfrac{1+i}{1-i} &= \dfrac{(1+i)(1+i)}{(1-i)(1+i)}\\ &= \dfrac{(1 + i)^2}{1 - i^2}\\ &= \dfrac{1 + 2i + i^2}{ 1 - (-1)}\\ &= \dfrac{2i}{2}\\ &= i\\ &= 0 + i \end{align*} This will be represented by a point equivalent of (0,1). Hence its argument will be $\dfrac{\pi}{2}$.\\

Correct Answer: C

Learning Objectives:

1. Represent a given complex number on an Argand diagram and vice versa.
2. Understand the three types of complex number representation- Cartesian, Polar and Exponential.
3. Understand the scenarios when representation in each form is suitable.
4. Represent a complex number in polar form given its modulus and argument.
5. Calculate the modulus and argument of a complex number written in Cartesian form.
6. Change between Cartesian, polar and exponential forms of a complex number.

1. Rectangular: Useful for addition and subtraction

\[ z = x + iy\]

2. Polar: Useful for multiplication and division

\[z = r (\cos \theta + i \sin \theta)\]

3. Exponential: Useful for powers and roots

\[z = r (e^{i \theta})\]

Solved Example: 9196-01

Represent the complex number: \[3 + 4i\] in its complex polar form \[z = r(\cos \theta + i \sin \theta)\]

A. r = 5, $\theta$ = 53.13$^\circ$

B. r = 0.2, $\theta$ = 53.13$^\circ$

C. r = 25, $\theta$ = 53.13$^\circ$

D. r = 5, $\theta$ = 36.86$^\circ$

\begin{align*} r &= \sqrt{(x^2 + y^2)}\\ r &= 5 \end{align*} \begin{align*} \theta &= \tan^{-1} \left(\dfrac{y}{x}\right)\\ \theta &= \tan^{-1} \left(\dfrac{4}{3}\right) = 53.13^\circ \end{align*}

Correct Answer: A

Solved Example: 9196-02

Let $z_1$, $z_2$ be the roots of the equation $z^2 - 6z + b = 0$ and $z_1$ and $z_2$ form an equilateral triangle with origin. Then, the value of b is:

A. 6

B. 12

C. 24

D. 36

Correct Answer: B

Learning Objectives: State and use the properties of algebraic operations (closure, commutativity, associativity, identity, inverse and distributivity) of complex numbers.

Addition and Subtraction of Complex Numbers

Add real part to real part
& imaginary part to imaginary part

Multiplication and Division of Complex Numbers

Multiply as if you multiplying a polynomial use \(j^2\) = -1

For division, multiply and divide by complex conjugate, so that the denominator will be a real number.

Solved Example: 1-11-03

Find the conjugate of $(6 + 5i)^2$.

A. $11 + 60i$

B. $11 - 60i$

C. $(6 - 5i)^2$

D. $\sqrt{(6 + 5i)}$

\begin{align*} (6 + 5i)^2 &= 36 +60 i+ 25i^2\\ &= 36 + 60 i + 25(-1)\\ &= 11 + 60i \end{align*} Complex conjugate of $11 + 60i$ is $11-60i$.

Correct Answer: B

Learning Objectives: Use De Moivre’s theorem to raise complex numbers to powers

\[\left(\cos \theta+i\sin \theta\right)^n=\cos(n\theta)+i\sin(n\theta)\]

Solved Example: 9930-01

What is \({\left[ {\dfrac{{\sin \dfrac{{\rm{\pi }}}{6} + {\rm{i}}\left( {1 - \cos \dfrac{{\rm{\pi }}}{6}} \right)}}{{\sin \dfrac{{\rm{\pi }}}{6} - {\rm{i}}\left( {1 - \cos \dfrac{{\rm{\pi }}}{6}} \right)}}} \right]^3}\) where \({\rm{i}} = \sqrt { - 1} ,\) equal to? (NDA Apr 2015 Math)

A. 1

B. i

C. -1

D. -i

Correct Answer: B

Solved Example: 9930-02

Evaluate: \[\left(\cos \dfrac{3\pi}{4} + i \sin \dfrac{3\pi}{4}\right)^8\]

A. -1

B. 1

C. 0

D. $-\pi$

\begin{align*} &\left(\cos \dfrac{3\pi}{4} + i \sin \dfrac{3\pi}{4}\right)^8\\ &=\left(\cos \dfrac{24\pi}{4} + i \sin \dfrac{24\pi}{4}\right)\\ &=\left(\cos 6\pi + i \sin 6\pi \right)\\ &=\left(1 + i (0) \right)\\ &= 1 \end{align*}

Correct Answer: B

Learning Objectives:

1. Find roots of complex numbers in polar form using De Moiver's theorem.

Euler’s identity

\[\begin{aligned} e^{j \theta} &= \cos \theta + j \sin \theta \\ e^{-j \theta} &= \cos \theta - j \sin \theta \end{aligned}\]

Roots of complex number

Take n\(^{th}\) root of the modulus and use Euler’s theorem to get multiple roots.

Solved Example: 1-13-01

Find the cube root of 8i lying in the first quadrant of the complex plane.

A. i-√3

B. 2i+√3

C. i+2√3

D. i+√3

Let z by the cube root of 8i. \[z^3 = 8i\] \[z^3 = 8(0 + i)\] \[z^3 = 8(\cos \dfrac{\pi}{2} + i \sin \dfrac{\pi}{2})\] Using De Moiver's Theorem, \[z_1 = 2(\cos (\dfrac{\pi}{6}) + i \sin (\dfrac{\pi}{6}))\] \[z_2 = 2(\cos (\dfrac{5\pi}{6}) + i \sin (\dfrac{5\pi}{6}))\] \[z_3 = 2(\cos (\dfrac{9\pi}{6}) + i \sin (\dfrac{9\pi}{6}))\] Out of this, $z_1$ is in the first quadrant. \begin{align*} z_1 &= 2(\cos (\dfrac{\pi}{6}) + i \sin (\dfrac{\pi}{6})) \\ &= 2(\dfrac{\sqrt{3}}{2} + i \dfrac{1}{2}) \\ &= \sqrt{3} + i (1) \end{align*}

Correct Answer: D

Solved Example: 1-13-02

The n$^{th}$ roots of any complex number are in:

A. Arithmetic progression

B. Geometric progression

C. Harmonic progression

D. No specific pattern

Correct Answer: B