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Solving Complex Number With Negative Fractional Exponent: i^(-21/2)

Asawira Emaan

New member
Nov 14, 2018
12
Kindly help me with this.
Solve
i^(-21/2)
Note: i means iota.
 

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MarkFL

Administrator
Staff member
Feb 24, 2012
13,735
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?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,687
Kindly help me with this.
Solve
i^(-21/2)
Note: i means 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:
$$(i)^{-\frac{21}{2}} = (e^{\frac\pi 2 i})^{-\frac{21}{2}}
= e^{-\frac{21\pi}{4}i} = \cos\left(-\frac{21\pi}{4}\right)+i\sin\left(-\frac{21\pi}{4}\right)
$$
 

Asawira Emaan

New member
Nov 14, 2018
12
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?
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:
$$(i)^{-\frac{21}{2}} = (e^{\frac\pi 2 i})^{-\frac{21}{2}}
= e^{-\frac{21\pi}{4}i} = \cos\left(-\frac{21\pi}{4}\right)+i\sin\left(-\frac{21\pi}{4}\right)
$$
but the answer of this question is -i (negative iota)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,687

MarkFL

Administrator
Staff member
Feb 24, 2012
13,735
but the answer of this question is -i (negative iota)
It is true that:

\(\displaystyle i^{-21}=-i\)

But:

\(\displaystyle i^{-\frac{21}{2}}\ne-i\)
 

Asawira Emaan

New member
Nov 14, 2018
12
Oh yes!
Thanks for help