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I have $M:=\sqrt{\frac{a\cdot(b+ic)}{de}}$ and all variables $a,b,c,d,e$ are real. Now I am looking for the real and imaginary part of this, but this square root makes it kind of hard.

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Which square root do you mean? –  Cameron Buie Aug 3 '13 at 18:04

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up vote 3 down vote accepted

$$\sqrt{\frac{a(b+ic)}{de}}=\sqrt{\frac{a}{de}}\cdot\sqrt{b+ic}$$

Let $$\sqrt{b+ic}=x+iy$$

$$\implies b+ic=(x+iy)^2=x^2-y^2+2xyi$$

Equating the real & the imaginary parts, $b=x^2-y^2, c=2xy$

So, $b^2+c^2=(x^2-y^2)^2+(2xy)^2=(x^2+y^2)^2\implies x^2+y^2=\sqrt{b^2+c^2}$

We have $$x^2-y^2=b$$

$$\implies 2x^2=\sqrt{b^2+c^2}+b\implies x^2=\frac{\sqrt{b^2+c^2}+b}2$$ $$\implies x=\pm\frac{\sqrt{\sqrt{b^2+c^2}+b}}{\sqrt2}$$

and $$\implies y^2=x^2-b=\frac{\sqrt{b^2+c^2}-b}2$$

$$\implies y=\pm\frac{\sqrt{\sqrt{b^2+c^2}-b}}{\sqrt2}$$

Now, the sign of $y=$ sign of $x\cdot$ sign of $c$

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@CameronBuie, please find the edited answer –  lab bhattacharjee Aug 3 '13 at 18:09

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