How do i solve for z for the following equation: z(1+i) = bar(z) + (3+2i). I know that conjugate of z is a-bi?

1 Answer
Feb 3, 2018

z=8-3i

Explanation:

Let z=a+bi, then barz=a-bi

and we can write z(1+i)=barz+(3+2i) as

(a+bi)(1+i)=(a-bi)+(3+2i)

or a+ai+bi+bi^2=(a-bi)+(3+2i)

or (a-b)+(a+b)i=(a-bi)+(3+2i)

and separating real and imaginary parts, we get

(a-b)+(a+b)i=(a+3)+(2-b)i

and as they have to be equal, we get

a-b=a+3 i.e. b=-3 and a+b=2-b i.e. a=2-2b=8

Hence z=8-3i