Question

Find & graph all the solutions in the complex plane?

Answer 1

\[\sqrt[5]{-1}\]

Answer 2

the following function is
\[i^{2/5}\]

Answer 3

How do I convert it to Euler form?

Answer 4

I feel like thats the first step perhaps?

Answer 5

haha yeah

Answer 6

Got any ideas? isn't it re^itheta, but what is the real part of this

Answer 7

never encountered with such power of iota :P

Answer 8

\[\sqrt[5]i=\sqrt[5]{-1^{1/2}}=(-1)^{1/10}\]

Answer 9

\[e^{i\pi}=-1\]

\[(-1)^{1/10}=(e^{i\pi})^{1/10}=e^{i\pi/10}\]

Answer 10

I still don't quite get it I found the real to be 1 now, and then theta = -90 from \[\tan^{-1} (-1/0) \]

Answer 11

putting it together I got \[((1)e^{i -\pi/4 + 2i \pi k})^5 \]

Answer 12

where k = 0,1,2.. but I didn't get your answer

Answer 13

none of the solutions are real

Answer 14

Ohhh I get what you did now!! Thanks for your help

Answer 15

\[\ddot\smile\]

Answer 16

can I ask u a quick question does that work for all the forms, what if you have a real like 1- i how would you change the form to e

Answer 17

would you have to do sqrt(1+1) = sqrt(2) to get the real?

Answer 18

sorry I meant like if you had sqrt(1-i)

Answer 19

im not sure about that one,

there should be only two roots though

Answer 20

Actually its alright I got it. How do you know what the graph looks like if you have them in e form?

Answer 21

@whpalmer4

Answer 22

http://www.wolframalpha.com/input/?i=x%5E2%3D1-i

Answer 23

Can't you just use De Moivre to take the square root after converting to Euler form?

Answer 24

Oh snap that's perfect didn't know of that theorem!

Answer 25

thank you so much! I have a test in an hour and didn't know how to do this thanks for all who replied

Answer 26

good luck with the test!