 # I am overthinking this but (-4)^{(-5/2)} is not a valid Jayleen Sanders 2022-01-22 Answered
I am overthinking this but ${\left(-4\right)}^{\left(-\frac{5}{2}\right)}$ is not a valid equation, am I correct? Working through the problem gives me $\frac{1}{-{4}^{\frac{5}{2}}}$ which then works out to $\frac{1}{\sqrt{-{4}^{5}}}$ which leaves a negative number in the square root, which is not valid.
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It is perfectly valid, the result is simply not a Real Number, it is what is called an Imaginary Number (or a Complex Number in the case in which you had a Real Numbered component as well), which you may not be familiar with.
The result you gave simplifies down to $-\frac{1}{32}i$, where $i$ is called the Imaginary Unit, and is equal to $\sqrt{-1}$
###### Not exactly what you’re looking for? gekraamdbk
The square root of a negative number is a complex number ${\left(-4\right)}^{-\frac{5}{2}}=\frac{1}{{\sqrt{-4}}^{5}}=\frac{1}{{\left(2i\right)}^{5}}=-\frac{1}{32}i$. If this sounds like complete gibberish to you, google "imaginary numbers" and read up on it.