The original problem

In an electricity course which I volunteered to help with, the students solve circuits using phasors. Using phasors requires a good knowledge of complex numbers arithmetics, because circuits are solved by expressing currents, voltages and impedances of resistors, inductors and capactors in the complex plane.

As part of this learning process, the students were asked to solve a very simple problem: $$\frac{2+2i}{1-2i}$$

The question asked specifically to solve this by converting numerator and denominator to polar form, because the students are meant to understand that the complex cartesian form is nice for adding and subtracting, while the polar form is nice for multiplying and dividing.

Whoever wrote this problem clearly wanted to make it as educational as possible, but only gave the solution: $-0.4 + 1.2i$

The naïve solution

So, if we stick to the question and convert numerator and deniminator to polar form, we get:

  • Convert $2+2i$ to polar form (step by step although it's trivial graphically):

    • $r=\sqrt{2^{2}+2^{2}}=2\sqrt{2}$
    • $\cos \phi = \frac{a}{r}=\frac{2}{2\sqrt{2}}=\frac{\sqrt{2}}{2}$
    • $\sin \phi = \frac{b}{r} = \frac{\sqrt{2}}{2}$
    • $\rightarrow \phi = \frac{\pi}{4}$
    • $\rightarrow2+2i = 2\sqrt{2}e^{i\frac{\pi}{4}}$
  • Convert $1-2i$ to polar form:

    • $r=\sqrt{1^{2}+2^{2}}=\sqrt{5}$
    • $\cos \phi = \frac{a}{r}=\frac{1}{\sqrt{5}}$
    • $\sin \phi = \frac{b}{r}=-\frac{2}{\sqrt{5}}$
    • $\tan \phi = -2$
    • $\rightarrow \phi = \tan^{-1}(-2)$
    • $\rightarrow 1-2i = \sqrt{5}e^{i \tan^{-1}(-2)}$
  • So the problem (for an educational example) starts here. $tan^{-1}(-2)$ can't be expressed as a fraction of $\pi$. Now, the division: $$\frac{2+2i}{1-2i} = \frac{2\sqrt{2}}{\sqrt{5}}\frac{e^{i\frac{\pi}{4}}}{e^{i \tan^{-1}(-2)}} = \frac{2\sqrt{2}}{\sqrt{5}}e^{i(\frac{\pi}{4}-\tan^{-1}(-2))}$$

  • And we convert the result back to cartesian form:

$$a = r \cos \phi = \frac{2\sqrt{2}}{\sqrt{5}} \cos (\frac{\pi}{4}-\tan^{-1}(-2)) = -0.4$$

$$b = r \cos \phi = \frac{2\sqrt{2}}{\sqrt{5}} \sin (\frac{\pi}{4}-\tan^{-1}(-2)) = 1.2$$

For the last step I had to resort to a computer, of course...

And so finally

$$\frac{2+2i}{1-2i} = -0.4+1.2i$$


Another way: complexe conjugates

Why not just multiply the original fraction by the complex conjugate of the denominator to get rid of the denominator's imaginary part, and you're left with a simple real division:

$$\frac{2+2i}{1-2i} = \frac{2+2i}{1-2i}\cdot\frac{1+2i}{1+2i} = \frac{2+2i+2i-4}{1+2i-2i+4} = \frac{-2+6i}{5} = -0.4+1.2i$$

How convenient!

My Question

So, after showing students both methods I also showed them how to solve $$\frac{2+2i}{1-i}$$ using the polar form, which is much easier and doesn't require a calculator. Then I got asked a very obvious question: which one should we use? Why do we teach them the polar form if the complexe conjugate is so much easier?

I had quite some trouble giving them a clear explanation.

Can you think of a division of complex numbers which is easier to solve using the polar form rather than the complexe conjugate? Only the really trivial ones like $\frac{2+2i}{1-i}$ which can be solved nearly mentally? :

$$\frac{2+2i}{1-i} = \frac{2\sqrt{2}}{\sqrt{2}}\cdot e^{i(\frac{\pi}{4}+\frac{\pi}{4})} = 2i$$

I feel like I'm missing something really obvious here...


I would say, as observed, the conjugation method for complex division is easier if the numbers are already given in $x+yi$ form. But, polar form may be more useful in questions like $\left|\displaystyle\frac{1+3i}{2-i}\right|=?$.

However, for instance, calculating a power of a complex number is easier using the polar form.

  • $\begingroup$ good point, i'll wait a bit if anyone has other comments. thanks! $\endgroup$ – Sébastien Dawans Mar 27 '13 at 16:08

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