Complex numbers written in exponential form.

z^4 = 81*e^(Pi/2)i

I have no idea how to find the roots for this equation. Can someone talk me through the algebra?

3*e^(pi/8 + n*pi/2)  for 0 <= n <= 3.

Just fourthroot the modulus and quarter the angle.

Here it is in a bit more detail.

Say z = s * e^it.
Then z^4 = (s * e^it)^4
         = s^4 * (e^it)^4
         = s^4 * e^4it

all by basic algebra.

Now we want z^4 = 81 * e^i(pi/2)
So we want to know s and t in
          s^4 * e^4it = 81 * e^i(pi/2)

Now the subtle step: we can decompose this into two equations:

          s^4 = 81
          e^4it = e^i(pi/2)

where s is a non-negative real and t is real in [0, 2pi).
Clearly, s = 3.  What about t?

It's tempting to say that 4t = pi/2, but this isn't necessarily correct, since e^ix is periodic with period 2pi.  Instead, we have:

          4t = (pi/2) + 2kpi    for some integer k

this gives us four essentially different solutions for t:

          t = pi/8
          t = pi/8 + pi/2
          t = pi/8 + pi
          t = pi/8 + 3pi/2

corresponding to k=0, 1, 2, or 3, respectively.

Since z = s*e^it, we have

          z = 3 e^i(pi/8)
          z = 3 e^i(pi/8 + pi/2)
          z = 3 e^i(pi/8 + pi)
          z = 3 e^i(pi/8 + 3pi/2)

And those are the four solutions.

I hope this was some help.

>>1
It is a mathematical rule that:
e^(xi) = cos(x) + i*sin(x)

>>4


Why are you bringing greek letters into it?

>>5
Fuck you. Then e^(xi) should read e^(x*i)

>>6
why are you bringing asterisks into it?

z^4 = 81*e^(Pi/2)i
z^4 = 81*e^((Pi/2)+2Kpi)i  (since you go round in a circle, provided k is an integer)
z = (81*e^((Pi/2)+2Kpi)i)^.25  (fourth root each side)
z = (81^.25)*e^(((pi/2)+2Kpi)i/4) (by laws of indicies, you can split and multiply the two exponents)
z=3*e^(((4k+1)pi)/8)i (rearrange)
z=3*e^(pi/8)i (k=0)
z=3*e^(5pi/8)i (k=1)
z=3*e^(11pi/8)i (k=2)
z=3*e^(15pi/8)i (k=3)
then cos+isin those angles to get them in cartesian form.

>>8
(4K+1)pi/8 = 11pi/8 when k = 2? What?

>>7 Sorry I forgot that I was removing the footnotes from my dissertation. I left that asterisk in by accident. Did I get rid of all the double-daggers, though? Good.

>>9
Whoops, numerical error. Yes I need to learn to count.
Other than that everythings right isn't it?

I wonder if OP that copied it straight of for his homework didn't notice that and got busted by his teacher...