# Polar form of a complex number

Question: Write the polar form of $$\frac{(1+i)^{13}}{(1-i)^7}$$

Well its obviously impractical to expand it and try and solve it. Multiplying the denominator by $(1+i)^7$ will simplify the denominator, and a single term in the numerator.

Answer I got: $$(\frac{1}{\sqrt2}(cos(\frac{\pi}{4}) + sin(\frac{\pi}{4})i)^{20}$$

Is this correct?

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I believe your answer is wrong. Just using the lengths of those numbers you get the length of the answer should be $(\sqrt2)^{13}/(\sqrt2)^7 = (\sqrt2)^6 = 8$. – DavidButlerUofA Jul 26 '14 at 20:03
Don't ever expand. Immediately convert $1+i$ and $1-i$ into polar form and go from there. – Lee Mosher Jul 26 '14 at 21:03

No, that's not correct. You must have made a couple of errors in your expansions. \begin{align} \frac{(1+i)^{13}}{(1-i)^7} &= \frac{(1+i)^{13}(1+i)^7}{(1-i)^7(1+i)^7} \\ &= \frac{1}{2^7}(1+i)^{20} \\ &= \frac{1}{2^7}\left(\sqrt{2}\left(\cos\frac{\pi}{4} + i \sin\frac{\pi}{4}\right)\right)^{20} \\ &= \frac{2^{10}}{2^7}\left(e^{i\pi/4}\right)^{20} \\ &= 8e^{5\pi i} \\ &= -8. \end{align} The polar form is $8(\cos\pi + i\sin\pi)$, or $(8,\pi)$.

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But that's not the polar form? Isn't polar form with $cos$ and $sin$?? – Gummy bears Jul 26 '14 at 20:14
True enough. If you wish, you can write it as $-8(\cos 0 + i\sin 0)$, or $8(\cos\pi + i\sin\pi)$. Either of those is a valid polar form. – rogerl Jul 26 '14 at 20:16
Distinction without a difference. – Vincent Jul 26 '14 at 20:19
Wait. I understood you up to where you have $cos\frac{\pi}{4}$ How do you get $cos\pi$ from that? – Gummy bears Jul 26 '14 at 20:19
@rogerl The polar form requires the parameter $r$ to be non-negative. – Git Gud Jul 26 '14 at 20:21

You can also convert numerator and denominator into polar form immediately to write

$$\frac{ [ \ \sqrt{2} \ cis(\frac{\pi}{4}) \ ]^{13} \ }{[ \ \sqrt{2} \ cis(-\frac{\pi}{4}) \ ]^7} \ \ .$$

DeMoivre's Theorem for powers gives us

$$= \ \frac{ (\sqrt{2})^{13} \ cis(\frac{13\pi}{4}) }{(\sqrt{2})^7 \ cis(-\frac{7\pi}{4})} \ \ .$$

Division of complex numbers in polar form then produces

$$= \ \frac{ (\sqrt{2})^{13} \ }{(\sqrt{2})^7} \ cis( \ \left[\frac{13\pi}{4} \right] \ - \ \left[-\frac{7\pi}{4} \right] \ ) \ \ .$$

You would simplify things from there. (Since the answer's already been posted, I'll finish this off:

$$= \ 2^{6/2} \ cis \left( \frac{20 \pi}{4} \right) \ = \ 2^3 \ cis(5 \pi) \ = \ 8 \ cis \ \pi \ \ \text{or} \ \ -8 \ \ . )$$

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(Slightly unrelated) I like $\mathrm{cis},$ pedagogically it just makes a lot of sense. – goblin Jul 26 '14 at 21:16
For some reason, many introductory texts don't use it, but most practitioners do. It's just a pain to write the factor out "longhand" in calculations with more than one line [you know the argument of both trig terms is the same!], especially in lecture, so I show it to students early on. (Of course, exponential notation is even quicker, but complex exponentials don't generally get presented even through two years of calculus courses...) – RecklessReckoner Jul 26 '14 at 21:18
I was thinking more along the lines of: its just a pain to try explaining why $e^{i\theta} = \cos \theta + i \sin \theta$ in a basic, introductory course in which the students have no analysis or knowledge of infinite summations under their belt. Better to just write $\mathrm{cis}\,\theta$ and be done with it. – goblin Jul 26 '14 at 21:20
DeMoivre's Theorem for powers. What is that? – Gummy bears Jul 27 '14 at 5:41
I keep getting stuck at $8(cos\frac{\pi}{4} + isin\frac{\pi}{4})^{20}$ What to do next? – Gummy bears Jul 27 '14 at 5:45