Problem involving square / square root of a complex number

[shillist]

Homework Statement
$z = (n + i)^{2}$

$n$ is a positive real number, and $arg(z) = \frac{\pi}{3}$

Find the value of $n$.

The attempt at a solution

I am reviewing old problem sets from years past, and came across this problem that appears pretty simple. I have my old answer as $n=\sqrt{3}$, which I can verify numerically as the correct answer. The problem is that I no longer remember the solution, and think I am missing a simple solution to this. We know the following:

$z^{1/2} = n + i$
$arg(n + i) = arctan(\frac{1}{n})$
$tan(arg(z)) = \sqrt{3}$

I think there must be some way to compare the arguments of the 2 sides and use the given fact that $arg(z) = \frac{\pi}{3}$ for a simple solution, but I don't see what it is. I am not sure how to deal with the square in the original problem statement, or the square root in my first equation.

 

[Dick]

If x is a complex number, how is arg(x) related to arg(x^2)? exp(it)^2=exp(2it), right?

 

[ehild]

Expand the right-hand side of the equation and compare the tangent of the arguments of both sides.

ehild

 

[shillist]

So in my case:
$arg(z^{1/2})=\frac{1}{2}arg(z)$
Then if we take $z^{1/2}=n+i$ and set the argument of each side equal, we have:
$arg(z^{1/2})=arg(n+i)$
$\frac{1}{2}\frac{\pi}{3}=arctan(\frac{1}{n})$
$\frac{\pi}{6} = arctan(\frac{1}{n})$
Taking the tangent of both sides:
$\frac{1}{\sqrt{3}}=\frac{1}{n}$
Thus:
$n=\sqrt{3}$

 

[Dick]

So in my case:
$arg(z^{1/2})=\frac{1}{2}arg(z)$
Then if we take $z^{1/2}=n+i$ and set the argument of each side equal, we have:
$arg(z^{1/2})=arg(n+i)$
$\frac{1}{2}\frac{\pi}{3}=arctan(\frac{1}{n})$
$\frac{\pi}{6} = arctan(\frac{1}{n})$
Taking the tangent of both sides:
$\frac{1}{\sqrt{3}}=\frac{1}{n}$
Thus:
$n=\sqrt{3}$

That's it. We can be a little bit sloppy about 2*pi factors in the arg because we know that 0<arg(1+n)<pi/2.

 

[shillist]

Thanks for the help!