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I am wondering how does

$$\frac{{{e^{zk}}}} {{{z^2} + 1}} = \frac{1} {{2i}}\left( {\frac{{{e^{zk}}}} {{z - i}} - \frac{{{e^{zk}}}} {{z + i}}} \right)?$$

I can see that $z^2 + 1 = (z + i)(z − i)$, but where does $\frac{1}{2i}$ come from?

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2 Answers 2

Note that for any $a,b\neq0$, $$\frac{1}{a}-\frac{1}{b}=\frac{b}{ab}-\frac{a}{ab}=\frac{b-a}{ab}.$$ Thus, with $a=z-i$ and $b=z+i$, $$\frac{1}{z-i}-\frac{1}{z+i}=\frac{(z+i)-(z-i)}{(z-i)(z+i)}=\frac{2i}{(z-i)(z+i)}=\frac{2i}{z^2+1}.$$ Thus $$\frac{1}{2i}\left(\frac{1}{z-i}-\frac{1}{z+i}\right)=\frac{1}{z^2+1},$$ and multiplying both sides by $e^{zk}$ we have that $$\frac{1}{2i}\left(\frac{e^{zk}}{z-i}-\frac{e^{zk}}{z+i}\right)=\frac{e^{zk}}{z^2+1}.$$

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Thanks, brilliant! –  Thomas Jan 19 '12 at 22:28
2  
@Thomas, for more generality, Google partial fraction. –  msh210 Jan 19 '12 at 22:29
    
Thanks, I just didnt realise I should use partial fractions to derive this. –  Thomas Jan 19 '12 at 22:40

$\displaystyle \frac{a}{z-i}+\frac{b}{z+i}=\frac{a(z+i)+b(z-i)}{z^2-i^2}=\frac{(a+b)z+(a-b)i}{z^2+1}$.

If you want this to be equal to $\displaystyle \frac{1}{z^2+1}$ for any $z$ you will need $a+b=0$ and $(a-b)i=1$.
I'll let you solve this system.

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Thanks, I just didnt realise I should use partial fractions to solve this! –  Thomas Jan 19 '12 at 22:40

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