Real part and imaginary part of a complex number of $i^{1/4}$

Question

Find the real and imaginary parts of $$i^{1/4}.$$

It seems to me that the real part is $0,$ because it does not appear and the imaginary part is $$1^{1/4},$$ but it seems too simple for it to be fine. Does the exponent affect anything?

Answer 1

If $z=a+ib,$ then $\Re z=a, \Im z=b.$ So you need to write $i^{1/4}$ in the form of $a+ib$ to find its real and imaginary parts.

Note that $$i=\cos(\frac{\pi}2+2n\pi)+i\sin(\frac{\pi}{2}+2n\pi)$$ so $$i^{1/4}=\cos(\frac{\pi}8+\frac{n\pi}2)+i\sin(\frac{\pi}8+\frac{n\pi}2).$$ So $$\Re i^{1/4}=\cos(\frac{\pi}8+\frac{n\pi}2), \Im i^{1/4}=\sin(\frac{\pi}8+\frac{n\pi}2),\quad n\in\{0,1,2,3\}.$$

Answer 2

Does $i^4=i$? I'd say that $i^4=1$. Note that $i=e^{\frac {i\pi} 2+2k\pi}$, so $$i^{1/4}=e^{\frac{i\pi}8+k\frac\pi 2}$$ for $k=0,1,2,3$.