Question

Complex numbers question!
Calculate this:

Answer 1

And represent it on a graph:
\[\sqrt[5]{1+i}\]

Answer 2

go there it may help okay

Answer 3

https://www.mathway.com/graph

Answer 4

think of it this way:
\(\sqrt[5]{1+i}=a\)
then \(a^5=1+i\) or
\(a^5=\sqrt2( \cos (\theta+n2\pi) +i sin(\theta +n2\pi))\) where \(\theta=45º \) and n=0,1,2,...\(\infty\)
from here, using Moovre's theorem:
\(a=\sqrt2(cos\frac{\theta+n2\pi}{5} +i sin\frac{\theta+n2\pi}{5})\)
now notice that for n=0,1,2,3,4 you will get different numbers, but startting from 5 and so on they will repeat. So there will be 5 different answers for a. Or to say it oyher way, 5 different roots

Answer 5

@myko Thank you very much! But can you explain to me once more why we five different root?

Answer 6

I would guess because a^5 = 1 + i has degree 5...just as every 5th degree polynomial has 5 roots, so would a^5 = 1 + i...

Answer 7

Ah ok, now I get it. Thank you everyone I really appreciate your help :)

Answer 8

in the last expretion for \(a\), just substitute n=0, n=1, n=2, n=3, n=4. You will get each time a different expretion. Each of which has argument \(2\pi/5\) bigger tan the previous one. After adding \(2\pi/5 \) five times you will come to same place that you started on the unit sircle. So the numbers will repeat for values higher than 4.

Answer 9

@naylah