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.

  • 1
    $\begingroup$ You could start by making the denominator a real number, then the numerator can be sorted out. Alternatively, you could make polar forms out of both numerator and denominator separately, then apply division rules to the results. $\endgroup$ – abiessu Sep 24 '16 at 21:50


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 $$\frac ND=\frac rs\mathrm e^{i(\theta-\varphi)}. $$


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

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

  • $\begingroup$ I seem to be missing something. How can the real part of this result be so small? The result should put the angle near the real line, and thus the real part should be larger than the imaginary part. Did I make a miscalculation? $\endgroup$ – abiessu Sep 24 '16 at 22:21
  • $\begingroup$ If I had to put it in the form Cos(x) + isin(x), how would I go about that? $\endgroup$ – cicero866 Sep 24 '16 at 22:23
  • $\begingroup$ @abiessu You were right, I wrote a wrong sign which made everything wrong. Please, check again and tell me if you find some other errors :) $\endgroup$ – Von Neumann Sep 24 '16 at 22:27
  • $\begingroup$ @cicero866 Just use the definition: $$e^{i\theta} = i\sin\theta + \cos\theta$$ Here $\theta = \arctan(\sqrt{3}-2)$ so it would be $$z = \sqrt{2}\left(i\sin\left( \arctan(\sqrt{3}-2) \right) + \cos \left (\arctan(\sqrt{3}-2)\right)\right)$$ $\endgroup$ – Von Neumann Sep 24 '16 at 22:28

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