# Thread: Argument of a complex number

1. ## Argument of a complex number

I've stuck on a basic exercise.
I must find all complex numbers $z$ such that $\arg (e^z)=\frac{\pi}{4}$.
Which is equivalent to say that $\arg (\underbrace{e^x \cos y}_{a} + i \underbrace {e^x \sin y}_{b} )= \frac{\pi}{4}$.

I remember a long time ago I've learned a formula where $a$, $b$ and $r$, the module of $z$ were involved that could solve for an argument of $z$. This is the only way that comes to my mind right now about how to solve the problem.

2. Originally Posted by arbolis
I've stuck on a basic exercise.
I must find all complex numbers $z$ such that $\arg (e^z)=\frac{\pi}{4}$.
Which is equivalent to say that $\arg (\underbrace{e^x \cos y}_{a} + i \underbrace {e^x \sin y}_{b} )= \frac{\pi}{4}$
Clearly you want $e^x\cos(y)=e^x\sin(y)>0$.
So any $x$ works; $y=\frac{\pi}{4}+2n\pi$.

Now can you explain these?

3. Originally Posted by Plato
Clearly you want $e^x\cos(y)=e^x\sin(y)>0$.
So any $x$ works; $y=\frac{\pi}{4}+2n\pi$.

Now can you explain these?
Thanks for the help.
Sure! We're looking for complex numbers in the first quadrant of the complex plane that are along the line $y=x$. Hence the real and imaginary parts must be equal and positive (due to the limitation of the first quadrant, because we have an argument of $\frac{\pi}{4}$).
This explain the first line of your post. To pass for the second line, divide by $e^x$ which is a positive number so there's no problem.
And now we must find $y$ such that $\cos y = \sin y>0$, which happens only when $y=\frac{\pi}{4}+2k\pi$, $k\in \mathbb{Z}$.

So the answer is $z=x+i (\frac{\pi}{4}+2k\pi)$, $k\in \mathbb{Z}$, $x\in \mathbb{R}$.