Splitting the square root of complex function into real and imaginary parts

Question

I have these functions below: $$\sqrt{(x+iy)^2-a^2}$$

$$\frac{b(x+iy)}{\sqrt{(x+iy)^2-a^2}}$$

How do I split these to get the real and imaginary parts of these functions?

If anyone could help me out, that really would be helpful!!!!

It will help me with the method of manufactured solutions for my fracture problem that has the Westergaard solution (as the analytical solution)!!!

Thank you very much, Mousumi

Thank you! I am very very new to stack exchange, and I am still learning. I apologize! — MousumiG, Dec 13 '17 at 08:58
No need to apologize. You are new around here and that's why I wrote what I wrote. — José Carlos Santos, Dec 13 '17 at 09:12

score 10 · Answer 1 · answered Dec 13 '17 at 09:05

Let $\sqrt{(x+iy)^2-a^2}=c+id$

$\implies (x+iy)^2-a^2=(c+id)^2$

Comparing the real & the imaginary parts

$\implies c^2-d^2=x^2-y^2-a^2\ \ \ \ (1)$

$2cd=2xy$

$(c^2+d^2)^2=(c^2-d^2)^2+(2cd)^2=(x^2-y^2-a^2)^2+(2xy)^2=(x^2+y^2)^2+a^4-2a^2(x^2-y^2)$

$c^2+d^2=\sqrt{(x^2+y^2)^2+a^4-2a^2(x^2-y^2)}\ \ \ \ (2)$

Use $(1),(2)$

Observe that $cd$ will have the same sign as $xy$

score 3 · Answer 2 · answered Dec 13 '17 at 09:07

3

Basically we need a complex number $A+ib$ such that its square equals $(x+iy)^2-a^2=(x^2-a^2-y^2)+i(2xy)$.

We have: $$(A+ib)^2=(A^2-b^2)+i(2Ab)=(x^2-a^2-y^2)+i(2xy)$$ giving us: $A^2-b^2=x^2-a^2-y^2$ and $Ab=xy$.

Now, we will get, $$A^2+b^2 = \sqrt{(A^2+b^2)^2} = \sqrt{(A^2-b^2)^2+4A^2b^2}=\sqrt{(x^2-a^2-y^2)^2+4x^2y^2}$$

Thus, $A^2=\frac{1}{2}[(x^2-a^2-y^2)+\sqrt{(x^2-a^2-y^2)^2+4x^2y^2}]$ and $b^2=\frac{1}{2}[\sqrt{(x^2-a^2-y^2)^2+4x^2y^2}-(x^2-a^2-y^2)]$.

Can you take it from here?

answered Dec 13 '17 at 09:07

Thank you, guys!!! I am still a bit confused with (c^2+d^2)^2 = (c^2-d^2)^2+(2cd)^2!!! But I will figure it out slowly! Thanks! – MousumiG Dec 13 '17 at 18:22
@MousumiG Note that $(c^2+d^2)^2=c^4+d^4+2c^2d^2= c^4 + d^4 - 2c^2d^2 + 4c^2 d^2 = , ?$ – Dec 14 '17 at 05:14

2 Answers2