Powers of Complex Numbers

[Phyisab****]

Homework Statement

Obtain z^10 for z=-1+i

Homework Equations

z=re^i(theta)

The Attempt at a Solution

Theta is 3pi/4. So z^10 = 32e^i(15pi/2).

The answer in my book is 32e^i(3pi/2) but I'm pretty sure that's wrong, can anyone confirm?

 

[latentcorpse]

yes but $e^{\frac{15 i \pi}{2}} = e^{\frac{12 i \pi}{2}} \cdot e^{\frac{3 i \pi}{2}}$

does that help?

 

[andylu224]

z = -1 + i = sqrt(2) cis(3/4 pi)

z^10 = 2^5 cis (10 x 3/4 pi) {De Moivre's Theorem}
= 32 cis (15/2 pi)
= -32i

 

[Phyisab****]

So I was right? Why did the answer appear in the book that way? Or is it wrong?

 

[andylu224]

Both the book and your answer is right -
though technically the answers should be 32e^-i(1/2pi).

remember, the range of principal argument is 0 < theta < pi (above the x-axis) &
-pi < theta < 0 (below the x-axis). so whenever you get a value beyond these ranges you have to convert your large value into an equivalent but smaller value within the range.

So you start at the line to the right of the origin of the x-axis (which is theta = 0). one revolution around all the quadrants back to the start is 2 pi. you keep on going until you finish up to 15/2 pi. you should end up at the line to the bottom of the origin of the y-axis. that is 3/2 pi.

 

[HallsofIvy]

yes but $e^{\frac{15 i \pi}{2}} = e^{\frac{12 i \pi}{2}} \cdot e^{\frac{3 i \pi}{2}}$

does that help?

So I was right? Why did the answer appear in the book that way? Or is it wrong?

The point is that $12 i \pi= 6 (2 i \pi)$ so that $e^{12 i\pi}= (e^{2 i\pi})^6= 1$.

$e^{15 i \pi}= e^{3 i\pi}$ but it is best always to write the argument between 0 and $2\pi$.