Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Which square root do you mean? –  Cameron Buie Aug 3 '13 at 18:04
add comment

1 Answer 1

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 $c$

share|improve this answer
    
@CameronBuie, please find the edited answer –  lab bhattacharjee Aug 3 '13 at 18:09
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.