Asawira Emaan said:Kindly help me with this.
Solve
i^(-21/2)
Note: i means iota.
MarkFL said:I think I would begin by writing:
$$z=i^{-\frac{21}{2}}=\left(\cis\left(\frac{\pi}{2}\right)\right)^{-\frac{21}{2}}=\cis\left(-\frac{21\pi}{4}\right)=\cis\left(\frac{3\pi}{4}\right)=\frac{-1+i}{\sqrt{2}}$$
Does that make sense?
Klaas van Aarsen said:Hi Asawira Emaan, welcome to MHB!
Are you aware that $i=e^{\frac\pi 2 i}=\cos(\pi/2)+i\sin(\pi/2)$?
It means that we can calculate it as:
$$(i)^{-\frac{21}{2}} = (e^{\frac\pi 2 i})^{-\frac{21}{2}}
= e^{-\frac{21\pi}{4}i} = \cos\left(-\frac{21\pi}{4}\right)+i\sin\left(-\frac{21\pi}{4}\right)
$$
Asawira Emaan said:but the answer of this question is -i (negative iota)
Asawira Emaan said:but the answer of this question is -i (negative iota)