Complex numbers roots

[palaszz]

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

Atm Im dealing with complex numbers, and Ive 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 Im 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!