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

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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?

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

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

Ray Vickson
Homework Helper
Dearly Missed
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?
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.

• FactChecker and Wi_N
nevermind. got it. thanks a lot.

Simon Bridge
Homework Helper
Great! What did you end up doing?

• Wi_N
Great! What did you end up doing?
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:
Ray Vickson
Homework Helper
Dearly Missed
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.

• 