Find the polar form of the complex number

Question

So I don't really understand how to find the polar form of a complex number. Especially when it has a numerator and denominator.

The complex number in question is $\dfrac{\sqrt{3} + i}{1 + i}$.

I know that when I want to find the $\operatorname{Im}{(z)}$ I would multiply everything by the conjugate, but I honestly have no idea where to even start for polar form.

edit: I need to put the answer in the form of something along the lines of sqrt(2)(cos(pie/integer) + isin(pie/integer)) I have no idea how they got there.

Answer 1

Hint:

Find the polar forms of the numerator and the denominator first: if $N=r\mathrm e^{i\theta}$ and $D=s\mathrm e^{i\varphi}$, then $$\frac ND=\frac rs\mathrm e^{i(\theta-\varphi)}. $$

Answer 2

First of all rationalise: multiply and divide the number you have by the conjugate of the denominator:

$$\frac{\sqrt{3} + i}{1+i}\cdot \frac{1-i}{1-i} = \frac{\sqrt{3}+i(-\sqrt{3} + 1) - 1}{2} = \frac{\sqrt{3} + 1}{2} + i\frac{1-\sqrt{3}}{2}$$

In this way you got your complex numebr in the form

$$z = A + iB$$

Now it's easy. The polar form is

$$z = |z|e^{i\theta}$$

$$|z| = \sqrt{A^2 + B^2} = \sqrt{\left(\frac{\sqrt{3}+1}{2}\right)^2 + \left(\frac{1-\sqrt{3}}{2}\right)^2} = \sqrt{2}$$

$$\theta = \arctan\left(\frac{B}{A}\right) = \arctan\left(\frac{1-\sqrt{3}}{\sqrt{3}+1}\right) = \arctan\left(\sqrt{3}-2\right)$$

So, unless you want to calculate the numeric value, you have:

$$\large z = \sqrt{2}\ e^{i\arctan(\sqrt{3}-2)}$$

Sine Cosine form

As wrote in the comment, to put in that form just use the definition:

$$e^{i\theta} = i\sin\theta + \cos\theta$$

Here $\theta = \arctan(\sqrt{3}-2)$ so it would be

$$z = \sqrt{2}\left(i\sin\left( \arctan(\sqrt{3}-2) \right) + \cos \left (\arctan(\sqrt{3}-2)\right)\right)$$