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.

#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

1. To find the complex numbers that satisfy the equation |z-1|=|z+3|=|z-i|, we can use the property of the modulus (absolute value) of a complex number.

Let's assume z = x + yi, where x and y are real numbers.

The equation |z-1| = |z+3| can be rewritten as |x + yi - 1| = |x + yi + 3|.

Simplifying this equation, we have:
√((x-1)^2 + y^2) = √((x+3)^2 + y^2).

Squaring both sides of the equation, we get:
(x-1)^2 + y^2 = (x+3)^2 + y^2.

Expanding and canceling the y^2 terms, we have:
(x^2 - 2x + 1) = (x^2 + 6x + 9).

Simplifying further, we get:
-2x + 1 = 6x + 9.

Bringing the x terms to one side and the constant terms to the other, we have:
-8x = 8.

Dividing by -8, we get:
x = -1.

Now, substituting x = -1 back into the equation (x-1)^2 + y^2 = (x+3)^2 + y^2, we have:
(1-1)^2 + y^2 = (-1+3)^2 + y^2.

Simplifying, we have:
0 = 4.

This equation is not true, which means there are no complex numbers that satisfy |z-1|=|z+3|=|z-i|.

Therefore, there are no solutions to this equation.

2. To find all cube roots of 8i, we can use the fact that the nth roots of a complex number can be found by converting the number to polar form and then using De Moivre's theorem.

Let's express 8i in polar form:
8i = 8(cos(π/2) + i*sin(π/2)).

Now, let's find the cube roots of 8i.

Using De Moivre’s theorem, we can raise 8i to the power of (1/3):
(8(cos(π/2) + i*sin(π/2)))^(1/3).

Using the property of exponents, we can distribute the exponent to both the real and imaginary parts:
8^(1/3) * (cos((π/2)/3) + i*sin((π/2)/3)).

Simplifying this expression, we have:
(2 * cos(π/6)) + i * (2 * sin(π/6)).

Using the values of sin(π/6) = 1/2 and cos(π/6) = sqrt(3)/2, we can further simplify:
(sqrt(3) + i) / 2.

Therefore, the three cube roots of 8i are:
(sqrt(3) + i) / 2,
(sqrt(3) + i) / 2,
and
(sqrt(3) + i) / 2.

Note that all three roots are the same since raising a complex number to the power of 1/3 yields three identical complex solutions.