# Finding Complex Solutions

• August 18th 2009, 12:18 AM
function
Finding Complex Solutions
Hey,
I'm not sure if i'm posting my question in the right forum, as it has to with complex numbers, but i will ask anyway and hopefully someone can help! I need to find all of the complex solutions for:
z^5 = i
Is anyone able to help me start this question?
Do i need to find i in polar form?

Thanks,
Function
• August 18th 2009, 12:45 AM
mr fantastic
Quote:

Originally Posted by function
Hey,
I'm not sure if i'm posting my question in the right forum, as it has to with complex numbers, but i will ask anyway and hopefully someone can help! I need to find all of the complex solutions for:
z^5 = i
Is anyone able to help me start this question?
Do i need to find i in polar form?

Thanks,
Function

Note that $z^5 = r^5 \text{cis} (5 \theta)$ and $i = \text{cis} \left( \frac{\pi}{2} + 2n \pi\right)$ where $n$ is an integer.

Edit: To save me some Moderating work, the following thread is probably relevant to what you're going to ask next: http://www.mathhelpforum.com/math-he...lex-roots.html. So don't ask it here ....
• August 21st 2009, 02:20 AM
function
Hey Mr. Fantastic,
Thanks for your help, although I'm still a little confused about how to approach this particular question. Once I have the i term, where do i go from there? Is there any particular formula that i need to use?

Thanks,
Function
• August 21st 2009, 02:30 AM
mr fantastic
Quote:

Originally Posted by function
Hey Mr. Fantastic,
Thanks for your help, although I'm still a little confused about how to approach this particular question. Once I have the i term, where do i go from there? Is there any particular formula that i need to use?

Thanks,
Function

You have $z^5 = r^5 \text{cis} (5 \theta) = \text{cis} \left( \frac{\pi}{2} + 2n \pi\right)$ where $n$ is an integer.

Therefore $z = r \text{cis} (\theta) = \text{cis} \left( \frac{\pi}{10} + \frac{2n \pi}{5} \right)$.

Now substitute five consecutive values of n to get the five distinct roots.
• August 22nd 2009, 01:59 AM
pacman
Roots in the complex plane,
z1 = i,
z2 = (-1)^(9/10),
z3 = -(-1)^(3/10),
z4 = -(-1)^(7/10),
z5 = (-1)(1/10)