Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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 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 at 21:03

2 Answers 2

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 $(8,\pi)$.

share|improve this answer
    
But that's not the polar form? Isn't polar form with $cos$ and $sin$?? –  Gummy bears Jul 26 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 at 20:16
1  
Distinction without a difference. –  Vincent Jul 26 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 at 20:19
2  
@rogerl The polar form requires the parameter $r$ to be non-negative. –  Git Gud Jul 26 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:

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

share|improve this answer
    
(Slightly unrelated) I like $\mathrm{cis},$ pedagogically it just makes a lot of sense. –  goblin Jul 26 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 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 at 21:20
    
DeMoivre's Theorem for powers. What is that? –  Gummy bears Jul 27 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 at 5:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.