1. Find (2+2i)^8 in Cartesian form
2. Square roots of -i
3. Show the following graphically |z+1|=|z-1| (abs value)
I am lost on 1 but my answer for 2 is i, sqrt[2](1-i)/2
as for 3, is it just a unit circle?
Thanks!
1) $\displaystyle (2+2i)^8 = ((2+2i)^2)^4 = (i8)^4=4096$.
2) Suppose $\displaystyle (a+ib)^2 = a^2-b^2 + 2iab = -i$. Equating real and imaginary parts gives |a|=|b| and ab=-1/2. The (two!) solutions are then $\displaystyle {\sqrt{2}\over2}(1-i)$ and $\displaystyle {\sqrt{2}\over2}(i-1)$.
3) This is the set of points z whose distance from 1 is the same as their distance from -1. The locus of points equidistant from two points is a straight line perpendicular to the line joining the two points and passing through its midpoint. In this case, it is the line on the imaginary axis.
It is indeed called conjugate. Observe that $\displaystyle \bar{z} := x-iy$ and $\displaystyle e^{i\theta} = \cos \theta + i \sin \theta$ so the conjugate of $\displaystyle e^{i\theta}$ is $\displaystyle \cos\theta - i \sin\theta=\cos\theta+i\sin(-\theta) = e^{-i\theta}$. So the conjugate of your z is $\displaystyle e^{2-ipi/4}$.