Complex Number Questions

1. Oct 28, 2014

teetar

Please, do not give me answers to these questions, I just want to have some hints/tips to point me in the correct direction, or if you're so inclined, you can use another similar/relevant to help me understand this concept. Thanks!
1. The problem statement, all variables and given/known data
a.)
Given that z/(z + 2) = 2-i, z ∈ ℂ, find z in the form a+bi
b.) Given that (a + bi)2 = 3 + 4i obtain a pair of simultaneous equations involving a and b. Hence find the two square roots of 3 + 4i.
2. Relevant equations
None, should be basic logic/algebra.

3. The attempt at a solution
a.)

z = (z + 2)(2 - i) = 2z - zi + 4 - 2i = 2z + 4 - zi - 2i
Is it then right to say that:
z = a + bi ∴ a = 2z + 4, b = -z - 2 ?
I could plug z into the original equation and try to do some long algebra:
2 - i = z/(z + 2) = (2z + 4 - zi - 2i)/(2z + 4 - zi - 2i + 2)
2 - i = (2z + 4 - zi - 2i) / (2z + 6 - zi - 2i)
2 - i = [2z + 4 - (z + 2)i] / [2z + 6 - (z - 2)i]
I could at this point multiply by the conjugate to get reals out of the denominator but that's a lot of work and I feel like I've already missed the point of this, as the question isn't worth too many points.
So, the most meaningful information I've presented relevant to this question (I think) is:
z = 2z + 4 - zi - 2i
2 - i = [2z + 4 - (z + 2)i] / [2z + 6 - (z - 2)i]

b.)
(a + bi)2 = (a + bi)(a + bi) = a2 + 2abi + b2
a2 + 2abi + b2 = 3 + 4i
a2 + b2 + 2abi = 3 + 4i
Hence, is it right to say that:
a2 + b2 = 3, 2ab = 4, so ab = 2 ?
This would mean that a = 2/b and likewise, b = 2/a
I can try substituting this into the other equation:
(2/b)2 + b2 = 3
(4/b2) + b2 = 3
4 + b4 = 3b2
4 = 3b2 - b4
Unfortunately, from here I'm completely lost, even though it's basic algebra, I can't quite put my finger on the next move. In fact, I'm not even sure if I've been doing anything remotely correct to this point.

2. Oct 28, 2014

Staff: Mentor

No. You have z = 2z -zi + 4 - 2i
Move all the z terms to one side, and solve algebraically for z.
You have a mistake right off the bat.
(bi)(bi) $\neq$ b2

3. Oct 28, 2014

RUber

For part b, you were on the right track by equating the real parts and imaginary parts. That is, assuming a and b are purely real.
Fix the sign error pointed out above and you should be in the clear.
If you again come to a point where you have a problem like $b^4+Cb^2+D=0$, remember to substitute $b^2=x$ and solve like any other quadratic.