Complex number solution (conjugate confusion): conjugateof(z) = 2*z + (1 - i)

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SUMMARY

The discussion centers on solving the equation conjugateof(z) = 2*z + (1 - i) for complex numbers. The user Mick attempts to express z as a + bi and the conjugate as a - bi, leading to the equation (a - bi) = 2*(a + bi) + (1 - i). The correct solution is identified as z = -1 + (1/3)i, achieved by equating the real and imaginary parts of the resulting equation -a - 3bi = 1 - i.

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mackhina
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Homework Statement



Find all complex solutions to:
conjugateof(z) = 2*z + (1 - i)

Homework Equations





The Attempt at a Solution



conjugateof(z) = 2*z + (1 - i)

I make:
z = (a + bi)
and
conjugateof(z) = (a - bi)

which gives:
(a - bi) = 2*(a + bi) + (1 - i)

then to find the roots:
0 = a + 3*bi + 1 - i
and that's where I get lost, I get the feeling I'm going in the right direction, but I don't seem to be able to reason out the right answer which in my book is:
z - -1 + (1/3)i

Any help would be heaps appreciated. Thanks in advance!

Cheers

Mick
 
Last edited:
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You have:

\overline{z}=2z+1-i

then letting z=a+bi:

-a-3bi=1-i

now equate real and imaginary parts on both sides of that equation.
 

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