Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $z=2-i$.

share|improve this question
4  
You can format your answers at MSE with $\LaTeX!$. Read basic manual at this meta post. –  FrenzY DT. Dec 4 '12 at 16:06
add comment

2 Answers

up vote 6 down vote accepted

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 $z = x+yi = 2-i$.

share|improve this answer
add comment

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 $3y-x=-5$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.