Question

(32 − 32i)^-6
how would i write this in polar form?

Answer 1

HI!!

Answer 2

I guess you first have to write \(32-32i\) in polar form right?

Answer 3

any idea how to do that? it is not too hard, you need two numbers \(r\) and \(\theta\)

Answer 4

i was wondering if my r would be \[\sqrt{2048}\]

Answer 5

and how i would turn that in \[\theta \]

Answer 6

wouldnt my theta be \[32/\sqrt{2048}+-32/\sqrt{2048}i\]

Answer 7

and turn that into cos and sin

Answer 8

@misty1212

Answer 9

Hey Art :)
Here is a neat little trick, it's probably not the way you're used to solving these though.\[\Large\rm \left(32-32\mathcal i\right)^{-6}\]Whenever I see that the numerical values are the same for the real and imaginary part, I immediately know that this is some type of pi/4 angle,
because sine and cosine give us the same value (maybe different sign) at pi/4, ya?


So what I want to do is, try to turn these 32's into sqrt2/2's.
Because then I can work backwards and turn my sqrt2/2's into some kind of cosine and sine of one of the pi/4 angles.

So I'll factor the 32 out of each term, and also another 2, so we end up with a 2 in the bottom. So a 64 in total.\[\Large\rm =64^{-6}\left(\frac{1}{2}-\frac{1}{2}\mathcal i\right)^{-6}\]I have to remember to apply my -6 whenever I pull something out.
I want to `gain` a sqrt2 in the top, so I'll divide out a \(\Large\rm \frac{1}{\sqrt2}\).\[\Large\rm =\frac{64^{-6}}{\sqrt2^{-6}}\left(\frac{\sqrt2}{2}-\frac{\sqrt2}{2}\mathcal i\right)^{-6}\]Let's get rid of the negative in that exponent out front,\[\Large\rm =\left(\frac{\sqrt2}{64}\right)^6\left(\color{orangered}{\frac{\sqrt2}{2}-\frac{\sqrt2}{2}\mathcal i}\right)^{-6}\]So umm, for this orange part...
Cosine corresponds to the real part,
sine the imaginary part, ya?

So we want to be in the quadrant where:
`cosine is positive`
`sine is negative`
Puts in quadrant 4.
So looks like we want 7pi/4.

Answer 10

\[\Large\rm =\left(\frac{\sqrt2}{64}\right)^6\left(\color{orangered}{\cos\left(\frac{7\pi}{4}\right)+\mathcal i\sin\left(\frac{7\pi}{4}\right)}\right)^{-6}\]

Answer 11

Then of course from here we can apply De'Moivre's Theorem! :)\[\Large\rm =\left(\frac{\sqrt2}{64}\right)^6\left(\color{orangered}{\cos\left(\frac{-6\cdot7\pi}{4}\right)+\mathcal i\sin\left(\frac{-6\cdot7\pi}{4}\right)}\right)\]

Answer 12

From there you just need to simplify your radius a bit,
and simplify your angle, unwind it.

Answer 13

Not the most typical way of doing this,
I apologize if it's too confusing :)

Answer 14

I understand somewhat but I see how you got everything and from there ill put the pieces together but that was a pretty sweet explanation.
Thanks!!!!!