# Complex Numbers in Exponential form

• May 23rd 2011, 02:21 AM
Blizzardy
Complex Numbers in Exponential form
Hey guys, got some questions which I need help in:

1) Express w = (10-2.(sq. root 3)i)/(1-3(sq.root 3)i) in the exponential form and hence or otherwise find the smallest +ve integer n such that w^n is a real number.

I expressed in exponential form and got w = 2e^i(\pi /3) but I got no idea how to do the 2nd part...

Another similar question which I need help in is this:

2)Given that z = (sq. root 3) + i, express z in the exponential form. Hence, find the real value of k such that z^5 + kz* = 0

Again, I only know how to express in exponential form... z = 2e^i(\pi /6) (Thinking)

• May 23rd 2011, 02:31 AM
Prove It
You should know that $\displaystyle e^{i\pi} = -1$. What power will you have to take $\displaystyle w$ to in order to get your exponent equal to $\displaystyle i\pi$?
• May 23rd 2011, 02:43 AM
Plato
Quote:

Originally Posted by Blizzardy
1) Express w = (10-2.(sq. root 3)i)/(1-3(sq.root 3)i) in the exponential form and hence or otherwise find the smallest +ve integer n such that w^n is a real number.
I expressed in exponential form and got w = 2e^i(\pi /3) but I got no idea how to do the 2nd part...

Clearly the answer to part 2 is 3.
Do you see why $\left( {2e^{\frac{\pi }{3}} } \right)^3 = - 2~?$
• May 23rd 2011, 03:19 AM
Blizzardy
Many thanks guys! Yupyup I understand Q1. =) But I still can't do Q2.

2)Given that z = (sq. root 3) + i, express z in the exponential form. Hence, find the real value of k such that z^5 + kz* = 0

So, z = 2e^i(\pi /6)
After substitution, I get: 2^5e^i(5pi/6) + ke^i(-pi/6) = 0
But how do I continue?
• May 23rd 2011, 04:07 AM
Plato
Quote:

Originally Posted by Blizzardy
Many thanks guys! So, z = 2e^i(\pi /6)
After substitution, I get: 2^5e^i(5pi/6) + ke^i(-pi/6) = 0
But how do I continue?

Just notice that $\exp \left( {\frac{{5\pi }}{6}} \right) = -\exp \left( {\frac{-\pi }{6}} \right)$.