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

• MHB
• Asawira Emaan
In summary, the conversation discusses solving for the value of i^(-21/2), with i representing iota. The expert suggests using the property that i=e^(pi/2 i) to calculate the value. The answer is determined to be (-1+i)/sqrt(2), not -i as mistakenly given.
Asawira Emaan
Kindly help me with this.
Solve
i^(-21/2)
Note: i means iota.

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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?

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:

$$\displaystyle z=i^{-\frac{21}{2}}=\left(\cis\left(\frac{\pi}{2}\right)\right)^{-\frac{21}{2}}=\cis\left(-\frac{21\pi}{4}\right)=\cis\left(\frac{3\pi}{4}\right)=\frac{-1+i}{\sqrt{2}}$$

Does that make sense?

Klaas van Aarsen 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:

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

But:

$$\displaystyle i^{-\frac{21}{2}}\ne-i$$

Oh yes!
Thanks for help

## 1. What is the meaning of a negative fractional exponent?

A negative fractional exponent represents the reciprocal or inverse of the base raised to the positive power of the fraction. In other words, it is the same as the base raised to the positive power of the fraction, but with the result being inverted.

## 2. How do you solve complex numbers with negative fractional exponents?

To solve complex numbers with negative fractional exponents, you can use the rules of exponents and the properties of complex numbers. First, rewrite the negative fractional exponent as a positive exponent using the reciprocal property. Then, apply the rules of exponents to simplify the expression. Finally, use the properties of complex numbers to express the answer in the form of a complex number.

## 3. Is it possible to have a negative fractional exponent for a complex number?

Yes, it is possible to have a negative fractional exponent for a complex number. This is because complex numbers can have both a real and an imaginary part, and the exponent can be applied to both parts separately. Therefore, the negative fractional exponent can be applied to the real or imaginary part of the complex number.

## 4. Can a negative fractional exponent result in a real number?

Yes, a negative fractional exponent can result in a real number. This can happen when the base is a negative number and the exponent is a rational number with an odd denominator. In this case, the result will be a real number because the negative number will be raised to an odd power, resulting in a negative value that can be simplified to a positive real number.

## 5. How can negative fractional exponents be used in real-world applications?

Negative fractional exponents are commonly used in engineering, physics, and other scientific fields to represent quantities that change over time or space. They are also used in financial calculations, such as compound interest, and in computer programming languages to represent complex numbers and perform calculations with them.

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