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!