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

AI Thread Summary
The discussion centers on solving the expression i^(-21/2), where i represents the imaginary unit. Participants clarify that i can be expressed in exponential form as e^(π/2 i), allowing for the calculation of i^(-21/2) using trigonometric identities. The correct solution is derived as (−1+i)/√2, while some participants mistakenly assert that the answer is simply -i. The conversation emphasizes the importance of understanding complex exponentiation and the distinction between different forms of the solution. Ultimately, the correct interpretation and calculation of the expression is confirmed.
Asawira Emaan
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Kindly help me with this.
Solve
i^(-21/2)
Note: i means iota.
 

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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?
 
Asawira Emaan said:
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:
$$(i)^{-\frac{21}{2}} = (e^{\frac\pi 2 i})^{-\frac{21}{2}}
= e^{-\frac{21\pi}{4}i} = \cos\left(-\frac{21\pi}{4}\right)+i\sin\left(-\frac{21\pi}{4}\right)
$$
 
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:
$$(i)^{-\frac{21}{2}} = (e^{\frac\pi 2 i})^{-\frac{21}{2}}
= e^{-\frac{21\pi}{4}i} = \cos\left(-\frac{21\pi}{4}\right)+i\sin\left(-\frac{21\pi}{4}\right)
$$

but the answer of this question is -i (negative iota)
 
Asawira Emaan said:
but the answer of this question is -i (negative iota)

It looks as if you were given the wrong answer then.
It's as MarkFL wrote, the answer is $\frac{-1+i}{\sqrt 2}$.
 
Asawira Emaan said:
but the answer of this question is -i (negative iota)

It is true that:

$$i^{-21}=-i$$

But:

$$i^{-\frac{21}{2}}\ne-i$$
 
Oh yes!
Thanks for help
 

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