How do I simplify the squareroot of -14i?

Question

I was thinking that I should separate the term $(-14i)^{0.5}$
into $(-14)^{0.5}$ and $i^{0.5}$
then into $i$ and $14^{0.5}$ and $i^{0.5}$

But Wolframalpha says that the answer is $(1-i)7^{0.5}$, and I don't think that's the same as $i^{1.5}14^{0.5}$.

How do I simplify $(-14i)^{0.5}?$

Answer 1

You're assuming that exponent laws like $(ab)^c = a^cb^c$ hold when dealing with complex bases and non-integer exponents. They don't. They hold as long as the bases are positive, real numbers. And they hold as long as the exponents are integers. But you need to fulfill at least one of those requirements for the laws to apply.

Also, I would be extremely catious in general when applying fractional exponents or roots to complex numbers. It is difficult to do correctly (it can be tricky to even define what "correctly" means).

How do I simplify $(-14i)^{0.5}?$

Hopefully, you stop writing things like $(-14i)^{0.5}$, and then you don't need to worry about how to calculate them any more. But if that's not an option for you, it often helps to think geometrically. Where in the complex plane is $-14i$? What geometric operation does $^{0.5}$ correspond to? Then just do some elementary geometry and you have an answer.

Answer 2

This separation rule is wrong with complex numbers. E.g. then we would have $$1 = ((-1) \cdot (-1))^{0.5} = (-1)^{0.5} \cdot (-1)^{0.5} = i \cdot i = -1$$

In general, pay attention to which exponentiation rules are still valid in $\mathbb{C}$.

I quite like to always use the following identites (which might even be definitions in your text book):

$$ \begin{align*} a^b& = \exp(b\ \mathrm{Log}(a)))\tag{1}\\ \mathrm{Log}(a)& = \log|a| + iArg(a)\tag{2} \end{align*} $$

By $\mathrm{Log}: \mathbb{C}^{\times} \to \mathbb{C}$ I denote the complex logarithm with branch cut $\mathbb{R}^{-}_0$. And by $\log: \mathbb{R}^+_{> 0} \to \mathbb{R}$ I denote the usual real logarithm.

Especially, using the first identity prevents me from assuming the exponential rules of the reals.

Let's apply them here:

$$ \begin{align*} &(-14i)^{0.5}\\ & = \exp(0.5\ \mathrm{Log}(-14i))\\ & = \exp(0.5\ (\log|-14i|+iArg(-14i)))\\ & = \exp(0.5\ (\log(14)-i\frac{\pi}{2}))\\ & = \exp(0.5 \log(14)) \exp(-i\frac{\pi}{4})\\ & = \sqrt{14} \exp(-i\frac{\pi}{4}) \end{align*} $$

That looks pretty neat, but we can do a little bit more. As per Kevin's comment, we can use

$$e^{-i \frac{\pi}{4}} = \frac{1-i}{\sqrt{2}}$$

You can either derive this with $\sin$/$\cos$ or use geometric reasoning: $-i\frac{\pi}{4}$ corresponds to $-45°$ ($\Rightarrow 1-i$) and you know that $|e^{i\varphi}| = 1$ for any $\varphi$, hence you just have normalize $1-i$ to get $\frac{1-i}{\sqrt{2}}$.

Applying this we get:

$$ \begin{align*} &(-14i)^{0.5}\\ & = \sqrt{7} (1-i) \end{align*} $$