The real part of z that satisfies i z^4+1=0 i | vedclass

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(B) Given,$i z^4 + 1 = 0$. $i z^4 = -1$. $z^4 = \frac{-1}{i} = i$. We know that $i = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2}$. So,$z^4 = \cos \frac{

Vedclass · Accepted Answer

$\cos \frac{\pi}{8}$

Vedclass · Answer

(B) Given,$i z^4 + 1 = 0$. $i z^4 = -1$. $z^4 = \frac{-1}{i} = i$. We know that $i = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2}$. So,$z^4 = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2}$. Using De Moivre's Theorem,$z = \left(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}\right)^{\frac{1}{4}} = \cos \frac{\pi}{8} + i \sin \frac{\pi}{8}$. Thus,the real part of $z$ is $\operatorname{Re}(z) = \cos \frac{\pi}{8}$.

The real part of $z$ that satisfies $i z^4+1=0$ is

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