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

Question

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

1. $z=-96+160i$
2. $z=96-160 i$
3. $z=160-96i$
4. $z=-160+96i$

what is the right answer ? Thanks in advance.

Answer 1

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

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

Answer 2

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 $Z=160i-96$