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I have the following expansions but I don't know how my teacher gets them. Apparently there is a formula for it (though the guy who told me didn't know it well), but I cannot find it in my notes.

For example, let $z$ be a complex number. Then,

$$\frac{1}{z^2+1}=\frac{i/2}{z+i}-\frac{i/2}{z-i}$$

It makes total sense to me in this case, but I get completely confused in this second case:

$$\frac{1}{(z^2+1)^2}=\frac{i/4}{z+i}-\frac{1/4}{(z+i)^2}-\frac{i/4}{z-i}-\frac{1/4}{(z-i)^2}$$.

What is/are the formulas that I need to compute this? Thanks!

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    $\begingroup$ For more general case treatment look at this $\endgroup$ – Math137 May 3 '14 at 20:36
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    $\begingroup$ This is splitting into partial fractions $\endgroup$ – vonbrand May 4 '14 at 3:28
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First you factor the bottom into: $$ (z^2 + 1)^2 = (z+i)^2 (z-i)^2. $$ Now for each factor that has a squared term, in this case both of them, you need to have one term of a variable over the factor and another term where the bottom has the factor squared. This is true in general. Therefore, you need to write it as $$ \dfrac{1}{(z^2+1)^2} = \dfrac{A}{z+i} + \dfrac{B}{(z+i)^2} + \dfrac{C}{z-i} + \dfrac{D}{(z-i)^2} $$ and then solve for $A$, $B$, $C$, and $D$ as you would before.

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    $\begingroup$ Thanks! I thought it would be a lot more complicated. I feel like an idiot now... :) $\endgroup$ – s1047857 May 3 '14 at 20:39
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    $\begingroup$ Hah, math is easy... once you see it. =D $\endgroup$ – Suugaku May 3 '14 at 20:48
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Start from the original equation $$ \frac{1}{z^2+1} = \frac{i/2}{z+i} - \frac{i/2}{z-i} $$ And just square it. $$ \frac{1}{(z^2+1)^2} = \left(\frac{i/2}{z+i}\right)^2 + \left(\frac{i/2}{z-i}\right)^2 - 2\frac{i/2}{z+i}\frac{i/2}{z-i} = -\frac{1/4}{(z+i)^2} - \frac{1/4}{(z-i)^2} + \frac{1/2}{z^2+1} $$ Reuse the first equation to get: $$ \frac{1}{(z^2+1)^2} = -\frac{1/4}{(z+i)^2} - \frac{1/4}{(z-i)^2} + \frac{i/4}{z+i} - \frac{i/4}{z-i} $$

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