Answer
Ans. $\frac{1}{√2}+i\frac{1}{√2}$, $\frac{1}{√2}+i\frac{1}{√2}$, $\frac{-1}{√2}+i\frac{-1}{√2}$, $\frac{-1}{√2}+i\frac{-1}{√2}$
Work Step by Step
Given:-$x^4+1=0$
$x^4=i^2$
$x^2=+i,-i$
$x=+i^{1/2}, -i^{1/2}$
We can write i as
i=cos(4k+1) π/2+isin(4k+1) π/2
$i^{1/2}=cos(4k+1) π/4+isin(4k+1) π/4$
Putting value of k=0, 1
We get
+($\frac{1}{√2}+i\frac{1}{√2}$), -($\frac{1}{√2}+i\frac{1}{√2}$)
Therefore
$x=+i^{1/2}, -i^{1/2}$
x have values
x=$\frac{1}{√2}+i\frac{1}{√2}$, $\frac{1}{√2}+i\frac{1}{√2}$, $\frac{-1}{√2}+i\frac{-1}{√2}$, $\frac{-1}{√2}+i\frac{-1}{√2}$
Ans.
