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

In summary, the conversation discusses solving a complex number equation and finding all the roots for which the modulus is less than 1. The process involves using Euler's equations and determining the values of x and y. The final solution is that the only root with a modulus less than 1 is z=0.
  • #1
palaszz
3
0
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?
 
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  • #2
Write z= x+ iy, with x and y real, and then [itex](\pi z)^2= \pi^2 (x^2- y^2+ 2xyi)[/itex]
So your equation becomes
[tex]e^{\pi^2(x^2- y^2+ 2xyi)}= e^{\pi^2(x^2- y^2)}e^{2\pi xyi}[/tex]

which has modulus [itex]e^{\pi^2(x^2- y^2)}[/itex].

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

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

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

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

Now, the condition that [itex]z= \sqrt{n}+ i\sqrt{n}[/itex] have modulus less than 1, requires that [itex]|z|= \sqrt{2n}< 1[/itex] which happens only for n= 0.
 
  • #3
HallsofIvy, sorry for the late reply
but I just wanted to thank you very much for your explanation.

It really helped a lot!
 

1. What are complex numbers roots?

Complex number roots are solutions to equations involving complex numbers, which are numbers that have both a real and imaginary component. They are often written in the form a + bi, where a is the real part and bi is the imaginary part (with i representing the square root of -1). The roots of a complex number can be found by setting the equation equal to 0 and solving for the complex variable.

2. How do I find the roots of a complex number?

To find the roots of a complex number, you can use the quadratic formula or the general formula for finding the roots of a polynomial. Simply plug in the values for the coefficients and solve for the complex variable. It is important to remember that complex numbers have two roots, known as the principal root and the conjugate root.

3. What is the principal root of a complex number?

The principal root of a complex number is the root with a positive imaginary part. In other words, it is the root that lies in the upper half of the complex plane. The principal root is often denoted as the "principal root" or "root 1" and is the first solution when using the quadratic formula.

4. What is the conjugate root of a complex number?

The conjugate root of a complex number is the root with a negative imaginary part. It is the mirror image of the principal root on the lower half of the complex plane. The conjugate root is often denoted as the "conjugate root" or "root 2" and is the second solution when using the quadratic formula.

5. Why are complex number roots important in mathematics and science?

Complex number roots are important in mathematics and science because they allow us to solve equations that involve imaginary numbers. They are commonly used in fields such as engineering, physics, and signal processing to model and understand real-world phenomena. Additionally, they have many applications in advanced mathematics, including complex analysis and number theory.

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