Mapping of complex exponential

[hadroneater]

Homework Statement

Determine the image of the line segment joining e^(i*2*pi/3) to -e^(-i*2*pi/3) under the mapping f = e^(1/2*Log(z)).

Homework Equations

The Attempt at a Solution

The line joining the two points: {z | -0.5 < x 0.5, y = sqrt(3)/2}
f = the principle branch of sqrt(z).

I am not sure what to do next.

 

[I like Serena]

Homework Statement

Determine the image of the line segment joining e^(i*2*pi/3) to -e^(-i*2*pi/3) under the mapping f = e^(1/2*Log(z)).

Homework Equations

The Attempt at a Solution

The line joining the two points: {z | -0.5 < x 0.5, y = sqrt(3)/2}
f = the principle branch of sqrt(z).

I am not sure what to do next.

Hi hadroneater!

So you have $z=x+i \frac 1 2 \sqrt 3$.
What would Log(z) be?

As an alternative path, that might be easier, suppose you write your line as $z=r e^{i\phi}$, with r a function of phi.
What would Log(z) in this case be?

 

[hadroneater]

So I have the image:
\left\{ e^{0.5Log(z)} | z = x + i√3/2}

z = \sqrt{x^{2} + \frac{3}{4}}*e^{i*arctan(\frac{2x}{sqrt{3})}

Log(z) = ln(\sqrt{x^{2} + \frac{3}{4}}) + i*arctan(\frac{2x}{sqrt{3})

How would finding Log(z) help me find the square root of z exactly?

Hmm...am I messing up the latex formatting?

 

[I like Serena]

You're missing a '}' in your latex.

Anyway, now that you have Log(z), can you multiply by (1/2), and take the exponential function?
What you will get, is indeed the (principle) square root of z.
TBH, I'm not sure what your problem is...?

Did you just want to find the square root of z?

 

[hadroneater]

Ah, I see. I just expected the answer to be in some neat closed form.

 

[I like Serena]

Ah well, what you have now contains a singularity in the arctan at x=0.
You could fix that by using for instance the arccot.

And you can reparametrize to the form $r(\phi)e^{i\phi}$, which will look better.
But your problem does not seem to ask for something like that...