Question

given z=√2-√2 i
how to solve z^13. i have no clue where to start

Answer 1

Express the complex number in polar form. If \(z=a+bi\), then
\(\large z=re^{i\theta}\)
where \(r=\sqrt{a^2+b^2}\) and \(\tan\theta =\dfrac{b}{a}\)

And then use DeMoivre's theorem, \(z^n =(re^{i\theta})^n=r^n(\cos(n\theta)+i\sin (n\theta))\)

Answer 2

could you perhaps show an example with the given numbers?, im not sure i understand it at all

Answer 3

\(z=\sqrt{2}-i\sqrt{2}\)
and \(z=re^{i\theta}\)

\(r=\sqrt{\left(\sqrt{2}\right)^2+\left(-\sqrt{2}\right)^2}=\sqrt{2+2}=2\)

\(\tan\theta=\dfrac{-\sqrt{2}}{\sqrt{2}}=-1\implies\theta= \arctan(-1)=\dfrac{-\pi}{4}\)

So... \(\large z=re^{i\theta}=2e^{-i\frac{\pi}{4}}\)

Recalling that \(e^{i\theta}=cos\theta+i\sin\theta\)... then

\(\large 2e^{-i\frac{\pi}{4}}=2\left[\cos\left(\frac{-\pi}{4}\right)+i\sin\left(\frac{-\pi}{4}\right)\right]\)

By DeMoivre's theorem: \(z^n =(re^{i\theta})^n=r^n(\cos(n\theta)+i\sin (n\theta))\), so:

\(\large \left(2e^{-i\frac{\pi}{4}}\right)^{13}=2^{13}\left[\cos\left(\frac{-13\pi}{4}\right)+i\sin\left(\frac{-13\pi}{4}\right)\right]\)

Now, you can convert the \(\dfrac{-13\pi}{4}\) into \(\dfrac{3\pi}{4}\) by adding 2 multiples of \(2\pi\)

Cn you proceed from here?

Answer 4

hmm im not sure, i already feel bad for nagging..but if you wouldnt mind it would be nice to have something to compare whatever answer i may get to.

Answer 5

im sitting on (-5792.62 + 5792,62i) as an answer

Answer 6

yes :), although I think it's preferable to use \(\sqrt{2}\) in your answer instead of putting it in decimal form ;)

Answer 7

hmm noted! thanks a lot for your time and effort in teaching me this!=) you saved my assignment;p

Answer 8

btw how do you come to that answer? i tried and it became a mess;p

Answer 9

the √2 answer that is

Answer 10

hm I suspect you just plugged that in your calculator lol.

Answer 11

Recall from the unit circle that \[\cos\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} \]
and
\[\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} \]

So:
To find \[ \cos\left(\frac{3\pi}{4}\right)\] and \[ \cos\left(\frac{3\pi}{4}\right)\], use symmetry of the unit circle: