Solving Complex Number Roots for e^(pi*z)^2 = i | Help Needed

[palaszz]

Hi out there peps, very nice forum! (my first topic)

Atm I am dealing with complex numbers, and I've got kinda problem solving this task. Hope for some help. Anyway, it sounds like this.

- Name all the roots for the equation e^((pi*z)^2)=i, for which modulus is less than 1.

Its obviously an exp. function, but I am unsure whether to use Eulers equations somehow?

 

[HallsofIvy]

Write z= x+ iy, with x and y real, and then $(\pi z)^2= \pi^2 (x^2- y^2+ 2xyi)$
So your equation becomes
$$e^{\pi^2(x^2- y^2+ 2xyi)}= e^{\pi^2(x^2- y^2)}e^{2\pi xyi}$$

which has modulus $e^{\pi^2(x^2- y^2)}$.

And, if that is to be equal to i we must have $e^{\pi (x^2- y^2)}= 1$ so that $x^2- y^2= 0$. That is, since x and y are real, x= y or x= -y.

We then have $e^{2\pi xyi}= cos(2\pi xy)+ i sin(2\pi xy)= i$ so that $cos(2\pi xy)= 0$ and $sin(2\pi xy)= 1$.

That happens when $2\pi xy$ is a multiple of $2\pi$ so that xy is equal to an integer, say n.

With x= y, that means $x^2= n$ so that $x= y= \sqrt{n}$. With x= -y, that means $-x^2= n$ which, since x is real, is impossible.

Now, the condition that $z= \sqrt{n}+ i\sqrt{n}$ have modulus less than 1, requires that $|z|= \sqrt{2n}< 1$ which happens only for n= 0.

 

[palaszz]

HallsofIvy, sorry for the late reply
but I just wanted to thank you very much for your explanation.

It really helped a lot!