# Thread: Complex equation

1. ## Complex equation

Hi I have this equation

$(e^{\pi\cdot z})^{2}=i$

Where I have to find the roots, when modulus er less than 1.

Cannot really solve it...

I have tried to $t=e^{\pi\cdot z}$

And then tried to solve it is a quacdric equation.

Where I get that

$e^{\pi x}\cos(\pi y)=1$ and $e^{\pi x}\sin(\pi y)=-2$

And also for the other root

$e^{\pi x}\cos(\pi y)=-1$ and $e^{\pi x}\sin(\pi y)=2$

But I have that $e^{\pi x}< 1 \Leftrightarrow x<0$

So anyone who could help me out or give me some tips?

2. Originally Posted by lo2
Hi I have this equation

$(e^{\pi\cdot z})^{2}=i$

Where I have to find the roots, when modulus er less than 1.

Cannot really solve it...

I have tried to $t=e^{\pi\cdot z}$

And then tried to solve it is a quacdric equation.

Where I get that

$e^{\pi x}\cos(\pi y)=1$ and $e^{\pi x}\sin(\pi y)=-2$

And also for the other root

$e^{\pi x}\cos(\pi y)=-1$ and $e^{\pi x}\sin(\pi y)=2$

But I have that $e^{\pi x}< 1 \Leftrightarrow x<0$

So anyone who could help me out or give me some tips?
$e^{2 \pi z} = e^{i\left(\frac{\pi}{2} + 2 n \pi\right)}$.

3. But is modulus not then 1 or more. Because x=0 and then we have.

$e^{0}(\cos(y)+i\sin(y))$

And thus modulus is 1.

4. following on from Mr F post.

$2 \pi z = i\left(\frac{\pi}{2} + 2 n \pi\right)$
$z = i\left(\frac{1+4n}{4}\right)$
$|z| = \left|\left(\frac{1+4n}{4}\right) \right|$

for what values of n is the modulus of that fraction less than 1 ?

Bobak

5. I am sorry but I do not even really understand the first calculation. How can
$(e^{\pi z})^{2}=e^{i(\frac{\pi}{2}+2n\pi)}$ And what is n and what happens to x.

Me trying:
$(e^{\pi z})^{2}=e^{2\pi z}=e^{2\pi (x+iy)}=e^{(2\pi x+2\pi iy)}$

6. Originally Posted by lo2
I am sorry but I do not even really understand the first calculation. How can
$(e^{\pi z})^{2}=e^{i(\frac{\pi}{2}+2n\pi)}$ And what is n and what happens to x.

Me trying:
$(e^{\pi z})^{2}=e^{2\pi z}=e^{2\pi (x+iy)}=e^{(2\pi x+2\pi iy)}$
Index Law: $(a^m)^n = a^{mn}$.

$i = e^{i \frac{\pi}{2}} = e^{i \left(\frac{\pi}{2} + 2 n \pi\right)}$. n is an integer: $n = 0, \, \pm 1, \, \pm 2, \, ....$

Why introduce z = x + iy etc. when you can solve for z directly?

7. Now I have finally managed to solve it and I get that.

$z=i(\frac{1}{2}+2)$

If I put in p=0 I get that $z=i \frac{1}{4}$

If I put p=1 then modulus is more than 1

But If I put in p=-1 then I get a modulus that is not too high and I get this solution $z=i -\frac{3}{4}$

But I am just not sure that it is in fact a solution to this equation. So is it?

8. $z = \frac{1}{4}i \ \ , \ \ - \frac{3}{4}i$ are all the solutions.

However I think you made a few typos in your pervious post.

Bobak

9. yeah...

But thank you very much! Your help is most appreciated!

10. Are you looking for solutions? I don't follow this thread.
$\left( {e^{\frac{{\pi i}}
{4}} } \right)^2 = i\,\& \,\left( {e^{\frac{{ - 3\pi i}}
{4}} } \right)^2 = i$

11. I just wanted to ask whether $z=-\frac{3}{4}i$ was a solution to the equation $(e^{z \pi})^2=i$

12. Originally Posted by lo2
I just wanted to ask whether $z=-\frac{3}{4}i$ was a solution to the equation $(e^{z \pi})^2=i$
Well I just wanted to ask you what $\Large {e^{\left( {\frac{{ - 3\pi i}}{4}} \right)}} =?$