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I am trying to calculate the residue of the function $$f(z)=\frac{2}{3z^2+8iz-3}$$ so as to evaluate the integral $$I=\int_{0}^{2\pi}\frac{1}{3\sin(\theta)+4}d\theta$$ I have found that $f$ has singularities at $z=\frac{(-4\pm\sqrt{7})i}{3}$ with only $z_0=\frac{(-4+\sqrt{7})i}{3}\in C_1(0)$ and that $z_0$ is a simple pole. I used the formula $$Res(f,z_0)=\frac{g(z_0)}{h'(z_0)}$$ since $z_0\in\mathbb{C}$ is a simple pole, $g,h$ are holomorphic on $D'_1(z_0)$, $h$ has a simple zero at $z_0$, and $g(z_0)\neq0$. Given the singularties of $f$, it can be written as $$f(z)=\frac{2}{\left(z-\frac{(-4+\sqrt{7})i}{3}\right)\left(z-\frac{(-4-\sqrt{7})i}{3}\right)}$$ However when I apply the above formula to find the residue, I find it to be $\frac{-6i}{\sqrt{7}}$ when I know the answer to be $\frac{-i}{\sqrt{7}}$. The answer scheme includes a factor of three as follows: $$Res(f,z_0)=\frac{2}{3\left(\frac{(\sqrt{7}+4)i}{3}+\frac{(\sqrt{7}-4)i}{3}\right)}$$ which gives the correct answer but I have no clue where this comes from. Am I missing something very simple here? Please respond using the above formula rather than the Laurent expansion etc. as I have several exercises similar to the above. Thanks in advance!

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  • $\begingroup$ I'm still looking through the question, but I will say starting from your initial integral my suggested method would be a Weierstrass sub, and I'm not exactly sure how you got the function we're taking the residues of $\endgroup$ Commented Apr 21, 2022 at 20:07
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    $\begingroup$ I think if you try multiplying out the denominator you give for $f(z)$ you'll find that's where the missing factor of $3$ is: basically, you forgot about the possibility of a constant factor. $\endgroup$ Commented Apr 21, 2022 at 20:14
  • $\begingroup$ Write your equation as $$f(z)=\frac23\frac1{z^2+8iz/3+1}$$What @StephenDonovan is saying that if you have $z_1,z_2$ the roots of the quadratic $az^2+bz+c=0$, then $$(z-z_1)(z-z_2)=z^2+\frac ba z+\frac ca$$ $\endgroup$
    – Andrei
    Commented Apr 21, 2022 at 20:16

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You may be overthinking this. Remember that the derivative of $az^2+bz+c$ is $2az+b$, so your formula with $g=2,h=3z^2+8iz-3$ gives

$\dfrac{2}{6z+8i}.$

All you need to do now is plug in the appropriate value $z=i(-4+\sqrt7)/3$ and simplify. Thus

$6z+8i=2i(-4+\sqrt7)+8i=2i\sqrt7$

$\dfrac{2}{6z+8i}=\dfrac{2}{2i\sqrt7}=\dfrac{-i}{\sqrt7}.$

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