Complex Equations with a complex number and its conjugate

**alexgeek** · September 13th 2010, 07:31 AM

Hi,
I'm wondering how to solve equations such as the following:
$2z - \bar{z}i = 1+4i$
I tried replacing z with (x+yi) and $\bar{x}$ with (x-yi) but that just gave me two unknowns.
My textbook has something similar in one of the exercises but no examples for it.
Thanks

**yeKciM** · September 13th 2010, 08:59 AM

Originally Posted by alexgeek

Hi,
I'm wondering how to solve equations such as the following:
$2z - \bar{z}i = 1+4i$
I tried replacing z with (x+yi) and $\bar{x}$ with (x-yi) but that just gave me two unknowns.
My textbook has something similar in one of the exercises but no examples for it.
Thanks

$2(x+iy) - (x-iy)i = 1 +4i$

$2x+2yi - xi - y = 1 +4i$

$(2x-y) +i(2y-x) = 1 +4i$

so complex numbers are equal if

$z_1 = x_1 + i y_1$

$z_2 = x_2 + i y_2$

$z_1 = z_2 \Rightarrow x_1 = x_2 \; \; and \; \; y_1 = y_2$

so you have ...

$2x+y = 1$

$2y-x = 4$

solve that (in any way that you know) and you will get your complex number Z = x+ iy

solution : $x= -\frac {2}{5} \;\; y= \frac {9}{5}$

**alexgeek** · September 13th 2010, 11:05 AM

Ahh I got that far and didn't see it like that. Sort of like equating coefficients.
Thanks

**yeKciM** · September 13th 2010, 11:13 AM

Originally Posted by alexgeek

Ahh I got that far and didn't see it like that. Sort of like equating coefficients.
Thanks

you're welcome

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