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I've just started calculating complex numbers (last time I calculated with complex numbers was in high school) and I've already got stuck at this exercise:

$$3z-i\bar z = 7-5i$$

where $\bar z$ is the conjugate of z.

What I've tested so far is to set $z=x+yi$

and with further calculations I've reached this equation

$$3(x+yi)-i(x-yi)=7-5i \implies 3x+3yi-xi+yi^2=7-5i$$

The result should be $z=2-i$.

Frenzy Li
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Rob
  • 83

4 Answers4

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You have as your last equation $$ 3x + 3yi - xi + yi^2 = 7-5i$$ now $i^2 = -1$, so we have $$ 3x-y + (3y - x)i = 7-5i$$ Now as $x$ and $y$ are real, we must have (as complex numbers are equal iff both real and imaginary parts are) $$ 3x-y = 7 \land 3y - x = -5 $$ This is a linear system for $x$ and $y$ which has $x = 2$ and $y = -1$ as its solution. Hence $z = x+yi = 2-i$.

martini
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A slightly different approach:

Start with the given equation

$3z-i\overline z = 7-5i$.

Any root of this equation must also solve the conjugate equation

$3\overline z-iz = 7+5i$.

The second equation forces $\overline z=[7+(z+5)i]/3$, so the original equation becomes by substitution

$3z-i[7+(z+5)i]/3 = 7-5i$.

This contains only $z$ and can be solved by the usual methods for linear equations to give $\color{blue}{z=2-i}$.

As a final step, check by direct evaluation that this root satisfies the original equation.

Oscar Lanzi
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If we know $i^2=-1$ and $x+iy=a+ib\iff x=a,y=b$

$$3x-y+i(3y-x)=7-5i$$

So, $3x-y=7$ and $3y-x=-5$

lab bhattacharjee
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[too long for a comment]

Since the complex number $ \ z \ = \ a + bi \ $ can be treated as the vector $ \ \langle a \ , \ b \rangle \ $ in the Argand plane, an equation of this sort is equivalent to a vector equation. We have the transformations $ \ \overline{z} \ \rightarrow \ \langle a \ , \ -b \rangle \ \ , \ $ $ \ iz \ \rightarrow \ \langle -b \ , \ a \rangle \ \ $ (90º CCW rotation), and $ \ -iz \ \rightarrow \ \langle b \ , \ -a \rangle \ \ $ (90º CW rotation). So $ \ 3z - i\overline{z} \ = \ 7 - 5i \ $ can be solved as $ \ 3 \langle a \ , \ b \rangle \ - \ \langle b \ , \ a \rangle \ = \ \langle 7 \ , \ -5 \rangle \ \ , \ $ using your preferred linear algebra technique.