$z=(-1+i)^{11}+(-1-i)^{15}=?$

Question

Can someone help me in this question : Let $z=(-1+i)^{11}+(-1-i)^{15}$ so

what is the right answer ? Thanks in advance.

You can use http://en.wikipedia.org/wiki/De_Moivre's_formula. — Cortizol, Jun 14 '14 at 11:35

Git Gud · Accepted Answer · 2014-06-14T14:46:25.690

1

Hint: $(-1-i)^2=-2i$ and $(-1+i)^2=2i$.

Alternatively write $-1-i=\sqrt{2}e^{5\pi i/4}$ and $-1+i=\sqrt 2e^{3\pi i/4}$.

edited Jun 14 '14 at 14:46

answered Jun 14 '14 at 11:34

Git Gud

score 1 · Answer 2 · answered Jun 14 '14 at 11:40

1

As said $(-1+i)^2=(-2i)$ and $(-1-i)^2=2i$. Hence $(-1-i)^{15}=(-1)(-128i)(1+i)=128i-128$

and $(-1+i)^{11}=(-32)(-1-i)=32i+32.$

Hence $Z=160i-96$

answered Jun 14 '14 at 11:40

2 Answers2