Solving Complex Quadratic Equations with Imaginary Coefficients

[Wi_N]

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

 

[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.

 

[Wi_N]

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

that doesn't give any answers.

 

[Ray Vickson]

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 I am 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.

 

[Wi_N]

nevermind. got it. thanks a lot.

 

[Simon Bridge]

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.

then just add. z=(p/2)+-(x+iy)

 

[Ray Vickson]

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.

then just add. z=(p/2)+-(x+iy)

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.

 

[Wi_N]

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.

no worries got the right answer. thanks.