Complex Function of z^(1/2): Find Solution

[pivoxa15]

Homework Statement

Find the complex function of z^(1/2))=(x+iy)^(1/2)

The Attempt at a Solution

The first step is z^(1/2)=e^((1/2)ln(z))=e^(1/2)[(ln|z|+i(theta)+2((pi)n)]

But the answers were not in this form.

 

[Gib Z]

What is your question asking? What do you mean by, find the complex function...

 

[CompuChip]

Do you want to extend the square root to the complex plane?
And what form were the answers in? Perhaps they are the same (just written down differently)?

 

[Kummer]

Homework Statement

Find the complex function of z^(1/2))=(x+iy)^(1/2)

The definition of $$a^b$$ for $$a,b\in \mathbb{C}$$ and $$a\not =0$$ is defined as $$\exp (b\ln a)$$.

So, $$z^{1/2} = \exp \left( \frac{\log z}{2} \right) = \exp \left( \frac{\ln |z|}{2} + i\cdot \frac{\arg z}{2} \right) = \sqrt{|z|}\cdot e^{i\arg(z)/2}$$

 

[pivoxa15]

Sorry, just to clarify the question is asking to find u(x,y) and v(x,y) where z^(1/2)=u(x,y)+iv(x,y) where z=x+iy.

 

[CompuChip]

That's called the real and imaginary part.
Recall Euler's identity
$$e^{i \phi} = \cos \phi + i \sin\phi.$$
Will that do?

 

[Kummer]

Sorry, just to clarify the question is asking to find u(x,y) and v(x,y) where z^(1/2)=u(x,y)+iv(x,y) where z=x+iy.

$$\sqrt{|z|}\cos \left( \frac{\arg z}{2} \right) + i \sqrt{|z|}\sin \left( \frac{\arg z}{2} \right)$$