Porisma

Worked solution

Solve an equation involving a conjugate

Solve the equation zz∗−2z+2z∗=5−4izz^*-2z+2z^*=5-4i.

CJC · Discussion · Question 3

Answer

z=2+iz=2+i or z=−2+iz=-2+i.

Expand and compare

Let z=a+biz=a+bi with a,b∈Ra,b\in\mathbb R. Then (a+bi)(a−bi)−2(a+bi)+2(a−bi)=5−4i(a+bi)(a-bi)-2(a+bi)+2(a-bi)=5-4i. Expand: a2−b2i2−2a−2bi+2a−2bi=5−4ia^2-b^2i^2-2a-2bi+2a-2bi=5-4i, hence a2+b2−4bi=5−4ia^2+b^2-4bi=5-4i. Imaginary parts give b=1b=1; real parts give a2+b2=5a^2+b^2=5, so a2=4a^2=4 and a=±2a=\pm2. Therefore z=2+iz=2+i or z=−2+iz=-2+i.

How did this part go?

Common mistakes

The conjugate reverses the imaginary part, not the real part.

Read the hint