# How do you write sqrt(d^11) as a exponential form?

Feb 1, 2016

d^(11/2

Feb 1, 2016

exponential form$= {d}^{\frac{11}{2}}$

#### Explanation:

Recall that the square root of a term can be written as $\sqrt{x}$ or ${x}^{\frac{1}{2}}$.

Thus:

$\sqrt{{d}^{11}}$

$= {\left({d}^{11}\right)}^{\frac{1}{2}}$

Multiply the exponents 11 and 1/2 together.

$= {d}^{\frac{11}{2}}$

$\therefore$, $\sqrt{{d}^{11}}$ in exponential form is ${d}^{\frac{11}{2}}$.

Feb 2, 2016

If $d$ is Real and non-negative then $\sqrt{{d}^{11}} = {d}^{\frac{11}{2}}$

Otherwise about the best you can say is $\sqrt{{d}^{11}} = {\left({d}^{11}\right)}^{\frac{1}{2}}$

#### Explanation:

Suppose $d = - 1$

Then ${d}^{11} = - 1$ and $\sqrt{{d}^{11}} = \sqrt{- 1} = i$

However, $d = \cos \left(\pi\right) + i \sin \left(\pi\right)$

So, using De Moivre's formula:

${d}^{\frac{11}{2}} = \cos \left(\frac{11 \pi}{2}\right) + i \sin \left(\frac{11 \pi}{2}\right)$

$= \cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right)$

$= 0 + i \left(- 1\right)$

$= - i$

So we find ${d}^{\frac{11}{2}} \ne {\left({d}^{11}\right)}^{\frac{1}{2}}$