 Thread starter
 #1
Asawira Emaan
New member
 Nov 14, 2018
 12
Kindly help me with this.
Solve
i^(21/2)
Note: i means iota.
Solve
i^(21/2)
Note: i means iota.
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Hi Asawira Emaan , welcome to MHB!Kindly help me with this.
Solve
i^(21/2)
Note: i means iota.
I think I would begin by writing:
\(\displaystyle z=i^{\frac{21}{2}}=\left(\cis\left(\frac{\pi}{2}\right)\right)^{\frac{21}{2}}=\cis\left(\frac{21\pi}{4}\right)=\cis\left(\frac{3\pi}{4}\right)=\frac{1+i}{\sqrt{2}}\)
Does that make sense?
but the answer of this question is i (negative iota)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:
$$(i)^{\frac{21}{2}} = (e^{\frac\pi 2 i})^{\frac{21}{2}}
= e^{\frac{21\pi}{4}i} = \cos\left(\frac{21\pi}{4}\right)+i\sin\left(\frac{21\pi}{4}\right)
$$
It looks as if you were given the wrong answer then.but the answer of this question is i (negative iota)
It is true that:but the answer of this question is i (negative iota)