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How do I decompose the fraction

$$\dfrac{1}{z^2 - 2i}$$

into partial fractions? I understand how to do partial fraction decomposition with real numbers, but I am unsure of how to do it with complex numbers. I attempted to find examples online, but all examples are with real numbers -- not complex.

I would greatly appreciate it if people could please take the time to demonstrate this.

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  • $\begingroup$ Factoring $\,z^2-2i\,$ would be a good starting point. $\endgroup$ – dxiv Aug 30 '17 at 22:02
  • $\begingroup$ @dxiv Yes...that is precisely what I am asking... $\endgroup$ – The Pointer Aug 30 '17 at 22:03
  • $\begingroup$ You do it in the same way over any field. In the case of complex numbers, it is simpler because you have only linear irreducible factors. $\endgroup$ – Bernard Aug 30 '17 at 22:04
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    $\begingroup$ $z^2-2i=(z+\sqrt{2i})(z-\sqrt{2i})=(z+1+i)(z-1-i)$. From here the process is identical to the decomposition of linear terms with real numbers $\endgroup$ – Alex Aug 30 '17 at 22:05
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    $\begingroup$ @ThePointer For the first equality, use difference of squares, $a^2-b^2=(a+b)(a-b)$. Next note that $\sqrt{2i}=\sqrt{2}(e^{i\pi/2})^{1/2}=\sqrt{2}e^{i\pi/4}=\sqrt{2} \frac{1}{\sqrt{2}}(1+i)=1+i$ $\endgroup$ – Alex Aug 30 '17 at 22:37
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Note that $z^2-2i=(z+\sqrt{2i})(z-\sqrt{2i})$ and $\sqrt{2i}=\sqrt{2}e^{i\pi/4}=1+i$. To simplify, let $b=1+i$, then $$\frac{1}{z^2-2i}=\frac{1}{(z+b)(z-b)}$$ From here it actually doesn't matter if you regard $b$ as real or complex, the process to find the partial fractions is the same as long as the terms are linear in $z$. So we let $$\frac{1}{(z+b)(z-b)}=\frac{A}{z+b}+\frac{B}{z-b}$$ for some $A,B\in \mathbb C$. Adding the two fractions on the right hand side we get that $$A(z-b)+B(z+b)=1$$ and so $$A+B=0$$ $$-bA+bB=1$$ which has solution $$A=-\frac{1}{2b}$$ $$B=\frac 1{2b}$$ Plugging in the original $b=1+i$ we have that $$\frac{1}{2b}=\frac12\frac 1{(1+i)}\frac{(1-i)}{(1-i)}=\frac 14(1-i)$$ Therefore $$\frac{1}{z^2-2i}=-\frac{\frac 14(1-i)}{z+1+i}+\frac{\frac 14(1-i)}{z-1-i}.$$ As you can see the process for computing the partial fraction coefficients with complex rationals is equivalent to that of real numbers.

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  • $\begingroup$ Thanks for the complete response. I did the calculations myself (with help from your post and those of others) and got $\dfrac{1}{(z - 1 - i)(z + 1 + i)} = \dfrac{1}{(z - 1 - i)(2 + 2i)} - \dfrac{1}{(z + 1 + i)(2 + 2i)}$. $\endgroup$ – The Pointer Aug 30 '17 at 23:30
  • $\begingroup$ For whatever reason, this answer absolutely captivates me! Stellar! $\endgroup$ – gen-z ready to perish Aug 30 '17 at 23:49
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the roots of $z^2 - 2i$ are $1+i$ and $-(1+i)$

$$ (1+i)^2 = 1^2 + 2 \cdot 1 \cdot i + i^2 = 1 + 2i - 1 = 2i $$

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Hint:  either will work to find the square roots of $\,2i\,$:

  • $\;\; 2i=(a+ib)^2 = a^2-b^2+2abi \implies a^2-b^2=0\,,\;2ab=2\,$

  • $\;\; 2 e^{i \pi/2} = \left(r e^{i \phi}\right)^2 \implies r^2 = 2\,,\;2\phi = \pi/2 + 2k\pi$

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I suppose one difficulty some will encounter here is how to factor $z^2-2i.$

You can observe that $$2i = 2(\cos90^\circ + i\sin90^\circ),$$ and therefore $$\pm\sqrt{2i\,} = \pm\sqrt 2(\cos45^\circ + i\sin45^\circ) = \pm \sqrt 2\left( \frac 1 {\sqrt 2} + i \frac 1 {\sqrt 2} \right) = \pm( 1+i).$$

Therefore $z^2 - 2i = \Big(z - (1+i)\Big)\Big( z+ ( 1+i)\Big).$

For the rest, if you know how to do other sorts of arithmetic with complex numbers, just proceed as with real numbers.

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