- #1
shillist
- 3
- 0
Homework Statement
[itex]z = (n + i)^{2}[/itex]
[itex]n[/itex] is a positive real number, and [itex]arg(z) = \frac{\pi}{3}[/itex]
Find the value of [itex]n[/itex].
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 [itex]n=\sqrt{3}[/itex], 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:
[itex]z^{1/2} = n + i[/itex]
[itex]arg(n + i) = arctan(\frac{1}{n})[/itex]
[itex]tan(arg(z)) = \sqrt{3}[/itex]
I think there must be some way to compare the arguments of the 2 sides and use the given fact that [itex]arg(z) = \frac{\pi}{3}[/itex] 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.
[itex]z = (n + i)^{2}[/itex]
[itex]n[/itex] is a positive real number, and [itex]arg(z) = \frac{\pi}{3}[/itex]
Find the value of [itex]n[/itex].
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 [itex]n=\sqrt{3}[/itex], 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:
[itex]z^{1/2} = n + i[/itex]
[itex]arg(n + i) = arctan(\frac{1}{n})[/itex]
[itex]tan(arg(z)) = \sqrt{3}[/itex]
I think there must be some way to compare the arguments of the 2 sides and use the given fact that [itex]arg(z) = \frac{\pi}{3}[/itex] 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.