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Problem.

Given a complex number $$z=2-2i$$

Find numbers $a$ and $b$ such that $$a+ib = \frac{1}{z}$$


I tried multiplying both sides by $z$ and got

$$(a+ib)(2-2i)$$ $$= 2a-2ai+2bi-2bi^2$$ $$=(2a+2b)-(2ai-2bi)=1$$

So I see that there is a real and complex part but I'm not sure how to continue. I've got 2 variables but only 1 equation. One solution would be $a=1/2$ and $b=0$.

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    $\begingroup$ Hint: What are the real and imaginary parts of the number $1$? $\endgroup$ – celtschk Apr 13 '14 at 13:24
  • $\begingroup$ BTW, with $a=1/2$ and $b=0$ the last equation would read $1-i=1$, which would mean $i=0$. Which quite obviously is not true. $\endgroup$ – celtschk Apr 13 '14 at 13:25
  • $\begingroup$ $2ai-2bi=0 \implies a=b$ and $2a+2b=1 \implies a=\frac{1}{4}$ $\endgroup$ – kingW3 Apr 13 '14 at 13:31
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You actually have two variables and two equations by comparing real and imaginary parts on both sides of your last equation. You should get you get $a=b$ since the imaginary part of $1$ is $0$ and $a=1/4$ after substituting $a=b$ into the equation you get from the real parts. This is the only solution.

The more standard method of doing this is to multiply $1/z$ on top and bottom by it's complex conjugate as illustrated in lab bhattacharjee's answer.

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$$\frac1{2-2i}=\frac12\cdot\frac1{1-i}=\frac12\cdot\frac{1+i}{1^2-(i)^2}=\frac{1+i}4$$

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