Thread: complex conjugates

Right, this comes in 2 parts, the 1st part is that this problem is part (b) of a question, in part (a) you are given z= 3+5i however i don't know if you can/need to use this fact for this part of the question (gotta love my teachers ). second part is the question itself:

Solve the equation z^2 + 2z*= -3, where z* denotes the complex conjugate of z.

thanks in advance for any help, really appreciate it.

Hello

Originally Posted by snowmonkey12

Right, this comes in 2 parts, the 1st part is that this problem is part (b) of a question, in part (a) you are given z= 3+5i however i don't know if you can/need to use this fact for this part of the question (gotta love my teachers ). second part is the question itself:

Solve the equation z^2 + 2z*= -3, where z* denotes the complex conjugate of z.

thanks in advance for any help, really appreciate it.

A possible approach is to work with z^2+2z*=-3 and with the conjugate of this equality :

This gives us two equations :



Adding and subtracting :



From (4) you can get the value of z+z*=2Re(z) which can then be used to get Im(z) from (3). Good luck !

Another way to do it:

To solve
z² + 2z* = -3

Let z = (x + iy); z* = (x - iy)

(x + iy)² + 2(x-iy) = -3
x² + 2xiy -y² + 2x - 2iy = -3

Equating Real and Imaginary parts:
Real: x² + 2x - y² = -3
Imaginary: 2xiy - 2iy = 0

From imaginary part:
2iy (x-1) = 0
So x = 1 or y =0

Substituting y = 0 into the real part:
x² + 2x + 3 = 0.
This quadratic has no real roots. We know that x has to be an integer, so this will not give us a solution.

Substituting x = 1 into the real part:
y² = 6
y= √6

Thus z = x + iy = 1 + √6i