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!

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- Apr 11th 2010, 10:20 PMJeffman503 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! - Apr 11th 2010, 10:33 PMmaddas
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. - Apr 11th 2010, 10:40 PMJeffman50
for 1), the problem says to find it in the form x+iy with x and y real.. 4096 is the answer straight up?

- Apr 11th 2010, 10:43 PMmaddas
Yes, with x=4096 and y=0 :]

- Apr 11th 2010, 10:43 PMJeffman50
ahh obviously!! thanks!

i hate these complex roots and stuff... too confusing - Apr 11th 2010, 10:45 PMJeffman50
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? - Apr 11th 2010, 11:21 PMJeffman50
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?) - Apr 11th 2010, 11:28 PMmaddas
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 $\displaystyle e^{2-ipi/4}$.