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$$\frac{z-a}{z+a}; a \in \Re$$

The part I'm confused about is the $a \in \Re$. I know that this means that $a$ is a real number (not imaginary), but then how do I interpret the addition/subtraction from $z$? Do I treat $z$ as $z=x+iy$? What is the concept to be learned here?

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The real number $a$ can be thought (and treated) as the complex number $a=a + i0 .$

With this in mind, operations like $z+a$, $z-a$ and so on, are just operations between complex numbers; for instance:

$$ z + a = (x + i y) + (a + i0 ) = (x+a) + i(y+0) = (x+a) + iy. $$

Now, to find the real and imaginary parts of $\frac{z-a}{z+a}$ you may start with

$$ \frac{z-a}{z+a} = \frac{z-a}{z+a} \cdot \frac{z-a}{z-a}= ...$$

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  • $\begingroup$ That's what my first attempt was but then I got stuck at $\frac{(z-a)^2}{x^2-y^2+2xyi+a^2}$. I'm not sure how to remove the $2xyi$. $\endgroup$ – whatwhatwhat Jan 25 '16 at 1:01
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You want something like difference of two squares, but so that the denominator ends up real. It turns out the right thing to do is to multiply by the complex conjugate of the denominator: $$ \frac{z-a}{z+a}\frac{z^*+a}{z^*+a} = \frac{\lvert z \rvert^2+a^2+2ia\Im(z)}{\lvert z \rvert^2 + 2a\Re(z)+a^2}, $$ which you can then easily split into real and imaginary parts, since everything but the $i$ is real.

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