Ray10
  • Ray10
Express \[z=8+8i\] in polar form and hence find \[z ^{8}\]? I've done the first part, just don't know how to do the second part
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Using de Moirves theorem thingy
anonymous
  • anonymous
I think it was something like:\[ [r(\cos(x)+i\sin(x))]^n = r^n(\cos(nx)+i\sin(nx)) \]
Ray10
  • Ray10
but they ask for the answer in polar form and I get a massive number :S can you please help me out with the theorem?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Ray10
  • Ray10
@wio could you show me how the theorem works please? I keep getting a ridiculous number :/
anonymous
  • anonymous
What do you have for \(x\) or I suppose it's usually \(\theta\)
anonymous
  • anonymous
What are \(r\) and \(\theta\)?
Ray10
  • Ray10
\[\theta \] comes up as \[\frac{ \Pi }{ 4 }\] r comes up as \[8\sqrt{2}\]
anonymous
  • anonymous
So what do you get for \(8\theta\) and for \(r^8\)?
Ray10
  • Ray10
\[r ^{8} = 128\] and \[8\theta \] = \[2\Pi \]
Ray10
  • Ray10
it asks for the polar form of; \[z ^{8} \]
anonymous
  • anonymous
Okay so you should be getting \[ 128(\cos(2\pi)+i\sin(2\pi)) = 128((1)+i(0)) = 128 \]
Ray10
  • Ray10
does that make the polar form; \[128CIS(2\Pi)\]
anonymous
  • anonymous
The polar form is just \(128\). The imaginary part is \(0\).
Ray10
  • Ray10
and the angle is what then?
anonymous
  • anonymous
The angle is \(2\pi = 0\).
anonymous
  • anonymous
|dw:1378451137190:dw|
anonymous
  • anonymous
They're not asking you to express \(z^8\) in polar form.
Ray10
  • Ray10
oh thank you @wio !! really appreciate it :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.