# Complex Form Conversions

🡑 Complex Numbers

## Cartesian to Polar

Using basic trigonometry, the following derivations can be derived

\begin{aligned} r &= \sqrt{a^2 + b^2} \\ \theta &= \arctan(\frac{b}{a}) \end{aligned}

## Polar to Cartesian

Going from polar to Cartesian can also be derived from trigonometry

\begin{aligned} a &= r \cos(\theta)\\ b &= r \sin(\theta)\\ \end{aligned}

## Examples

### Convert to Polar Practice – First Quadrant

Convert z = 1 + j\sqrt{3} from Cartesian to polar form.

#### solution

\begin{aligned} a &= Re(z) = 1 \\ b &= Im(z) = \sqrt{3} \\ \\ r &= \sqrt{a^2+b^2} \\ &= \sqrt{1^2+\sqrt{3}^2} \\ &= 2 \\ \\ \theta &= \arctan(\frac{b}{a}) \\ &= \arctan(\frac{\sqrt{3}}{1}) \\ &= \frac{\pi}{3} = 60 \degree \\ \end{aligned} \begin{aligned} \therefore z &=1 + j\sqrt{3} \\ &=2e^{i\frac{\pi}{3}} = 2\angle{\frac{\pi}{3}} \\ &=2e^{i60\degree} = 2\angle{60\degree} \end{aligned}

### Convert to Cartesian Practice – First Quadrant

Convert z = 2\angle{\frac{\pi}{6}} = 2\angle{30\degree} from Cartesian to polar form.

#### solution

\begin{aligned} r &= 2 \\ \theta &= \frac{\pi}{6}\\ \\ a &= r \cos(\theta)\\ &= 2 \cos(\frac{\pi}{6})\\ &= \sqrt{3}\\ \\ b &= r \sin(\theta)\\ &= 2 \sin(\frac{\pi}{6})\\ &= 1\\ \end{aligned} \begin{aligned} \therefore z &=2\angle{\frac{\pi}{6}} \\ &=\sqrt{3}+j \\ \end{aligned}

### Convert to Polar Practice – Second Quadrant

Convert z = -\sqrt{3} + j from Cartesian to polar form.

Hint: Pay attention to the complex angles

#### solution

\begin{aligned} a &= Re(z) = -\sqrt{3} \\ b &= Im(z) = 1 \\ \\ r &= \sqrt{a^2+b^2} \\ &= \sqrt{(-\sqrt{3})^2+1^2} \\ &= 2 \\ \\ \theta &= \arctan(\frac{b}{a}) \\ &= \arctan(\frac{1}{-\sqrt{3}}) \\ &= -\frac{\pi}{6} = -30 \degree \\ \end{aligned}

Given the fact that a is negative, and b is positive, we know that the complex number must be located somewhere in the second quadrant. Thus \theta = -\frac{\pi}{6} = -30 \degree is not logical as it would put our complex number somewhere in the fourth quadrant. Due to the periodic nature of the tan function, we can add \pm \pi or \pm 180\degree for an alternate angle.

I chose to add \pi to keep the angle [-\pi,\pi] due to personal preference.

\begin{aligned} \theta &= \pi + \theta \\ \theta &= \pi + (-\frac{\pi}{6}) \\ \theta &= \frac{5\pi}{6} \\ \end{aligned}

As expected, this angle brings our complex number into the second quadrant.

\begin{aligned} \therefore z &=-\sqrt{3} + j \\ &=2e^{i\frac{5\pi}{6}} = 2\angle{\frac{5\pi}{6}} \\ &=2e^{i150\degree} = 2\angle{150\degree} \end{aligned}

### Convert to Cartesian Practice – Second Quadrant

Convert z = 2\angle{\frac{2\pi}{3}} = 2\angle{120\degree} from Cartesian to polar form.

#### solution

\begin{aligned} r &= 2 \\ \theta &= \frac{2\pi}{3}\\ \\ a &= r \cos(\theta)\\ &= 2 \cos(\frac{2\pi}{3})\\ &= -1\\ \\ b &= r \sin(\theta)\\ &= 2 \sin(\frac{2\pi}{3})\\ &= \sqrt{3}\\ \end{aligned} \begin{aligned} \therefore z &=2\angle{\frac{2\pi}{3}} \\ &=-1+j\sqrt{3} \\ \end{aligned}

### Convert to Polar Practice – Third Quadrant

Convert z = -1 – j\sqrt{3} from Cartesian to polar form.

Hint: Pay attention to the complex angles

#### solution

\begin{aligned} a &= Re(z) = -1 \\ b &= Im(z) = -\sqrt{3} \\ \\ r &= \sqrt{a^2+b^2} \\ &= \sqrt{(-1)^2+(-\sqrt{3})^2} \\ &= 2 \\ \\ \theta &= \arctan(\frac{b}{a}) \\ &= \arctan(\frac{-\sqrt{3}}{-1}) \\ &= \frac{\pi}{3} = 60 \degree \\ \end{aligned}

Given the fact that both a is negative, and b is negative, it is clear that the complex number should be in the third quadrant. Thus \theta = \frac{\pi}{3} = 60 \degree is wrong as it would put our complex number somewhere in the first quadrant. Due to the periodic nature of the tan function, we can add \pm \pi or \pm 180\degree for an alternate angle.

I added -\pi to keep the angle between [-\pi,\pi] due to personal preference.

\begin{aligned} \theta &= -\pi + \theta \\ \theta &= -\pi + (\frac{\pi}{3}) \\ \theta &= -\frac{2\pi}{3} \\ \end{aligned}

As expected, this angle brings our complex number into the third quadrant.

\begin{aligned} \therefore z &=-1 – j\sqrt{3} \\ &=2e^{i-\frac{2\pi}{3}} = 2\angle{-\frac{2\pi}{3}} \\ &=2e^{i-120\degree} = 2\angle{-120\degree} \end{aligned}

### Convert to Cartesian Practice – Third Quadrant

Convert z = 2\angle{-\frac{5\pi}{6}} = 2\angle{-150\degree} from Cartesian to polar form.

#### solution

\begin{aligned} r &= 2 \\ \theta &= -\frac{5\pi}{6}\\ \\ a &= r \cos(\theta)\\ &= 2 \cos(-\frac{5\pi}{6})\\ &= -\sqrt{3}\\ \\ b &= r \sin(\theta)\\ &= 2 \sin(-\frac{5\pi}{6})\\ &= -1\\ \end{aligned} \begin{aligned} \therefore z &=2\angle{-\frac{5\pi}{6}} \\ &=-\sqrt{3}-j \\ \end{aligned}

### Convert to Polar Practice – Fourth Quadrant

Convert z = \sqrt{3} – j from Cartesian to polar form.

#### solution

\begin{aligned} a &= Re(z) = \sqrt{3} \\ b &= Im(z) = -1 \\ \\ r &= \sqrt{a^2+b^2} \\ &= \sqrt{\sqrt{3}^2+(-1)^2} \\ &= 2 \\ \\ \theta &= \arctan(\frac{b}{a}) \\ &= \arctan(\frac{-1}{\sqrt{3}}) \\ &= -\frac{\pi}{6} = -30 \degree \\ \end{aligned}

As expected, this angle brings our complex number into the second quadrant.

\begin{aligned} \therefore z &=\sqrt{3} – j \\ &=2e^{i\frac{\pi}{6}} = 2\angle{\frac{\pi}{6}} \\ &=2e^{i-30\degree} = 2\angle{-30\degree} \end{aligned}

### Convert to Cartesian Practice – Fourth Quadrant

Convert z = 2\angle{-\frac{\pi}{3}} = 2\angle{-60\degree} from Cartesian to polar form.

#### solution

\begin{aligned} r &= 2 \\ \theta &= -\frac{\pi}{3}\\ \\ a &= r \cos(\theta)\\ &= 2 \cos(-\frac{\pi}{3})\\ &= 1\\ \\ b &= r \sin(\theta)\\ &= 2 \sin(-\frac{\pi}{3})\\ &= -\sqrt{3}\\ \end{aligned} \begin{aligned} \therefore z &=2\angle{-\frac{\pi}{3}} \\ &=1-j\sqrt{3} \\ \end{aligned}