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!