# 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 $\displaystyle z$ such that $\displaystyle \arg (e^z)=\frac{\pi}{4}$.
Which is equivalent to say that $\displaystyle \arg (\underbrace{e^x \cos y}_{a} + i \underbrace {e^x \sin y}_{b} )= \frac{\pi}{4}$.

I remember a long time ago I've learned a formula where $\displaystyle a$, $\displaystyle b$ and $\displaystyle r$, the module of $\displaystyle z$ were involved that could solve for an argument of $\displaystyle z$. 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 $\displaystyle z$ such that $\displaystyle \arg (e^z)=\frac{\pi}{4}$.
Which is equivalent to say that $\displaystyle \arg (\underbrace{e^x \cos y}_{a} + i \underbrace {e^x \sin y}_{b} )= \frac{\pi}{4}$
Clearly you want $\displaystyle e^x\cos(y)=e^x\sin(y)>0$.
So any $\displaystyle x$ works; $\displaystyle y=\frac{\pi}{4}+2n\pi$.

Now can you explain these?

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

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 $\displaystyle y=x$. 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 $\displaystyle \frac{\pi}{4}$).
This explain the first line of your post. To pass for the second line, divide by $\displaystyle e^x$ which is a positive number so there's no problem.
And now we must find $\displaystyle y$ such that $\displaystyle \cos y = \sin y>0$, which happens only when $\displaystyle y=\frac{\pi}{4}+2k\pi$, $\displaystyle k\in \mathbb{Z}$.

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