Question

How do I find the value of i^(1/4)?

Answer 1

I know you have to turn it to polar form

Answer 2

x^1/4 is the same as d fourth root of X

Answer 3

i dont get what your saying @diamondboy

Answer 4

\[i=\cos(\frac{\pi}{2}+2n \pi)+i \sin(\frac{\pi}{2}+2n \pi) \text{ where } n \text{ is an integer } \]

Answer 5

now raise both sides to 1/4 power

Answer 6

what do i put for n?

Answer 7

how did you get the theta?

Answer 8

n=0,1,2,3
you will give the 4 answers you seek
I just recalled that cos(pi/2)=0 and sin(pi/2)=1

Answer 9

i=a+bi
when a=0 and b=1 right?

Answer 10

and cos is 0 when theta=pi/2
and sin is 1 when theta=pi/2

Answer 11

we never had to do that, so i dont want to do something different

Answer 12

are u talking about complex numbers here @_alex_urena_

Answer 13

let z=i^1/4
\[z^4=i=r(\cos \frac{ \pi }{ 2 }+i \sin \frac{ \pi }{ 2 })=r e ^{i \frac{ \pi }{ 2 }}=re ^{i \left( \frac{ \pi }{ 2 }+2n \pi \right)}\]
\[z^4=r e ^{i\left( 4n+1 \right)\frac{ \pi }{ 2 }}\]
here r=1
\[z=e ^{i \left( 4n+1 \right)\frac{ \pi }{ 8 }},n=0,1,2,3\]
?

Answer 14

yass, powers and roots of complex numbers

Answer 15

we never did something like that @surjithayer

Answer 16

ok so what if u did this