# Linear algebra - find all solutions with complex numbers

(a)Find all t $$\epsilon C$$ such that $$t^{2}$$ + 3t + (3-i) = 0. Express your solution(s) in teh form x+iy where x,y $$\epsilon R.$$

(b) Prove that | 1+iz | = | 1-iz | if and only if z is real.

Okay so I tried to use the quadratic formula to find the roots to find the solutions, but I am stuck because I have a complex number within the square roots.

t = -b +/- sqrt(b^2 - 4ac) / 2a
t = -3 +/- sqrt[(-3)^2 - 4(1)(3-i)] / 2(1)
t = -3 +/- sqrt(-3 + 4i) / 2

what do i do?

also for question (b), where do I even start?

## Answers and Replies

Homework Helper
1. Well, you should know how to take square roots of imaginary numbers, if you're supposed to use that method. Since you do not, it seems, why not retry the question without using the quadratic formula which you probably weren't supposed to use anyway. For example, I hope you wouldn't use the quadratic formula on x^2 - 5x +6 to find the roots of 2 and 3.

2. Iz z=x+iy, you want to show y=0. Well, what does the condition |1-iz|=|1+iz| imply?