Question

1. Find all complex numbers z such that |z-1|=|z+3|=|z-i|, if i is the square root of -1.

I got to the part where you graph the equations, but got stuck.

2. Find all cube roots of 8i.

I know there are 3 answers, but I am stuck.

Answer 1

#1.
|z-1|=|z+3|
|z-i|=|z+3|
are both straight lines.

This can be seen since
|z-(a+bi)|=|z-(c+di)|
|(x+yi)-(a+bi)|=|(x+yi)-(c+di)|
|(x-a)+(y-b)i|=|(x-c)+(y-d)i|
(x-a)^2 + (y-b)^2 = (x-c)^2 + (y-d)^2
x^2-2ax+a^2+y^2-2by+b^2 = x^2-2cx+c^2+y^2-2dy+d^2
(2c-2a)x + (2d-2b)y = c^2+d^2-a^2-b^2
Now just plug in your numbers and plot the lines for their intersection.

#2.
8i = 2^3 cis(?/2)
so, the main cube root is
2cis(?/6)
Now add multiples of 2?/3
see the plot at

http://www.wolframalpha.com/input/?i=x%5E3%3D8i