3

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 $(-14i)^{0.5}?$

helpme
  • 661

2 Answers2

5

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:

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

ComFreek
  • 1,489
  • 1
    Well ok you are right (it didn't simplify too much) but you could also write $$e^{-\pi i/4}=\cos\left(\frac{-\pi}{4}\right)+i \sin\left(\frac{-\pi}{4}\right) = \frac{1-i}{\sqrt{2}}$$ with Principal value taken. –  Oct 01 '19 at 13:05
  • Ooh it simplifies to Wolframalpha's answer – helpme Oct 01 '19 at 13:57
5

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.

Arthur
  • 199,419
  • 2
    +1 for the geometry. If you take my solution $\sqrt{14} \exp(-i\frac{\pi}{4})$ and square it, then you see that you'll square the length (and get length $14$) and the angle (and get $-i$). So in total you get $-14i$! – ComFreek Oct 01 '19 at 13:05
  • The argument of $(-14i)^{0.5}$ should be $-pi/4$, but mine is $3pi/4$. It seems that I am "rotating" anticlockwise by $3pi/4$ twice instead of clockwise $-pi/4$ twice to obtain -14i (on the negative side of the imaginary axis). This has led to me obtaining $(i-1)7^{0.5}$ instead of $(1-i)7^{0.5}$. Why should I be "rotating" clockwise instead of anticlockwise? Isn't anticlockwise the convention? Sorry if this seems a bit unclear. – helpme Oct 01 '19 at 13:21
  • @helpme That's entirely understandable. Note how I said "it can be tricky to even define what "correctly" means" above? This is what I meant. One common standard is to rotate away from the negative real axis when taking square roots. If you don't know this, then you will get the "wrong" answer some times. But the fact of the matter is that until you impose some arbitrary, artificial limits like that, there are two answers. And using an expression like $(-14i)^{0.5}$ and then having it have two values at the same time is problematic. – Arthur Oct 01 '19 at 13:23
  • (continued) There are two solutions to this problem. One is to impose these arbitrary, artificial limits like "square roots rotate away from the ngative real axis". The other possibility is to just accept that an expression like $(-14i)^{0.5}$ is ill-defined and should be avoided. I like to go for the second solution, but many insist that the first solution works just fine. And to be fair, it does. Most of the time. – Arthur Oct 01 '19 at 13:24
  • What happens if one somehow obtains $(−14i)^{0.5}$ when trying to solve an equation? Is that a sign of an error in the working? – helpme Oct 01 '19 at 13:47
  • 1
    @helpme No. It's a sign that you have taken a completely sensible equation like $x^2 = -14i$, and then applied the square root to it. And that's sensible as well, as long as you know to be very careful when doing it. It is not an actual error, but it's something that I personally would go to great lengths to avoid because it's very easy to make a mistake somewhere down the line. – Arthur Oct 01 '19 at 13:59