# Solving z^2+pz+qiz+a+bi=0

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1. Feb 17, 2017

### Wi_N

• Thread moved from a technical forum, so homework template missing
ok having major problems. i can easily solve z^2 + pz +a+bi=0 solutions but that extra qiz is really annoying me.

z^2 + 3z+4iz-1+5i=0

(z+2i)^2+3z-5+5i=0
z+2i = w, z=w-2i

w=-3(w-2i)+5-5i

then im not getting anything sensible for solving x and yi. what am i doing wrong?

2. Feb 17, 2017

### Simon Bridge

The real and imaginary parts need to be zero at the same time ... simultaneous equations.
If z is complex, then put z = a+ib.

3. Feb 17, 2017

### Wi_N

4. Feb 17, 2017

### Ray Vickson

Just use the quadratic formula on the equation $z^2 + (3+4i) z +(5i-1) = 0.$ This has the form $z^2 + bz + c=0$, with $b = 3+4i$ and $c = 5i-1$. The only real complication is that the discriminant $b^2 - 4 a c$ can be complex---it equals $-3 + 4i$ in your case---so you need to be able to take square roots of complex numbers.

5. Feb 17, 2017

### Wi_N

nevermind. got it. thanks a lot.

6. Feb 17, 2017

### Simon Bridge

Great! What did you end up doing?

7. Feb 18, 2017

### Wi_N

basically took the root out of ((p/2)^2-q)^(1/2)

(p/2)^2-q = x+iy

(a+bi)(a+bi)=(x+iy)

a^2-b^2=x (if x is positive a^2>b^2 if not b^2>a^2)
2abi=iy = > 2b=y (if y is positive then a and b both must either be negative or positive, if y is negative then either x or y has to be negative or vice versa and the other positive)

this way you can figure out your roots which will always be 2 roots since its a z^2 equation.

Last edited: Feb 18, 2017
8. Feb 18, 2017

### Ray Vickson

Please write down the final solutions (to your numerical example) explicitly. We cannot tell if you have made an error if you will not show us what you got.

9. Feb 20, 2017

### Wi_N

no worries got the right answer. thanks.