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What is the imaginary part of this function? $$\displaystyle\frac{-2\mu(x)-2\mu(iy)}{x^{2}+2xiy-y^{2}+b^{2}}$$

I need just the imaginary part for the equation of streamlines in a fluids question but I don't know how to get it as there is an i in the denominator.

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Rationalizing the denominator by multiplying with the conjugate would work. Are both $\mu(x)$ and $\mu(iy)$ real? – Argon Feb 18 '13 at 21:11

2 Answers 2

Hint: Assuming $\mu$ is a real function and $b^2\in \Bbb R$, multiply by $\displaystyle 1=\frac{(x^2-y^2+b^2)-2xyi}{(x^2-y^2+b^2)-2xyi}$

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you meant $y^2$ I presume. Thanks – Amzoti Feb 18 '13 at 21:15
@Amzoti Thank you. – Git Gud Feb 18 '13 at 21:15

Hint: Try computing

$$\frac{-2\mu(x)-2\mu(iy)}{x^{2}+2xiy-y^{2}+b^{2}}\cdot \frac{(x^2-y^2+b^2)-2xyi}{(x^2-y^2+b^2)-2xyi}$$

$$ = \frac {-2\mu(x)-2\mu(iy)}{(x^2 - y^2 + b^2) + 2xyi} \cdot \;\frac{(x^2-y^2+b^2)-2xyi}{(x^2-y^2+b^2)-2xyi}$$

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Then we multiply the conjugates in the denominator, which rids us of $i$ in the denominator. – amWhy Feb 18 '13 at 21:20

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