1. 3 Complex Number questions

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!

2. 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.

3. for 1), the problem says to find it in the form x+iy with x and y real.. 4096 is the answer straight up?

4. Yes, with x=4096 and y=0 :]

5. ahh obviously!! thanks!

i hate these complex roots and stuff... too confusing

6. wait the way i tried to do b didnt work.. where did i go wrong?

-i in polar coordinates is: e^(i3pi/2)

so -1= cos(3pi/2) + isin (3pi/2)
and the the other root is = cos(5pi/2) + isin (5pi/2)

so then you divide each angle by 2 right? and then solve?

7. hmm.. one more problem i can't figure out

Let z= e^(2+ipi/4)

what is the conjugate of z? (I htink its called a conjugate.. where there is a dash on top of the z?)

8. 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 $\displaystyle e^{2-ipi/4}$.