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I am trying to find $i^{-1}$. I already know that the answer is $-i$, but I can't figure out a way to determine that using math. This is what I am doing: $$i^{-1}$$ $$\frac1i$$ $$\sqrt{\left(\frac1i\right)^{2}}$$ $$\sqrt{\frac{1}{-1}}$$ $$\sqrt{-1}$$ $$\pm i$$ What am I doing wrong?

user224474
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    Multipñy and divide by $i$. The square root is only defined for nonnegative reals. – chubakueno Mar 18 '15 at 01:16
  • The property $\sqrt{ab}=\sqrt{a}\sqrt{b}$ does not work for general complex $a$ and $b$. Consider, for example, $\sqrt{1}=\sqrt{(-1)^2} "=" (\sqrt{-1})^2 = i^2 = -1 $. – Chappers Mar 18 '15 at 01:17

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Do this. $$i(-i) = -i^2 = -(-1) = 1.$$

ncmathsadist
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I'm not sure where your 3rd expression comes from. In general, to find $\frac{1}{a+bi}$, where $a,b\in \mathbb{R}$, multiply numerator and denominator by the conjugate $\overline{a+bi}=a-bi$ to obtain $$\frac{1}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2+b^2}$$ In your case, take $a=0,b=1$ to obtain $\frac{1}{i}=\frac{-1i}{1}=-i$.

The Substitute
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The law $\sqrt{ab} = \sqrt{a}\sqrt{b}$ doesn't hold for complex numbers, else we would have $$ i = \sqrt{-1} = \sqrt{\frac{1}{-1}} = \frac{\sqrt{1}}{\sqrt{-1}} = \frac{1}{i} = -i. $$

Mathmo1123
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You asked “what am I doing wrong”, and nobody has answered that yet.

$\sqrt{x^2}$ is not in general equal to $x$. That identity is true only for positive real numbers. Consider $x=-3$ for example: $\sqrt{(-3)^2} = \sqrt 9 = 3 \ne -3$.

In fact, $\sqrt x$ is not even defined except when $x$ is a positive real number.

So when you passed from $\frac1i$ to $\sqrt{\left(\frac1i\right)^2}$, what you did was invalid. $\frac1i$ means one thing, and $\sqrt{\left(\frac1i\right)^2}$ means nothing at all.

MJD
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  • Generally, $\sqrt x$, when $x$ is negative, is defined to be $i\sqrt{-x}$. (Of course, $-i\sqrt{-x}$ would also have been a perfectly fine choice.) – Akiva Weinberger Mar 18 '15 at 01:28
  • Generally, it is not. – MJD Mar 18 '15 at 01:30
  • How often have you seen equations stating $\sqrt{-1}=i$ and $\sqrt{-3}=i\sqrt3$ compared to, say, $\sqrt{-1}=-i$ and $\sqrt{-3}=-i\sqrt3$? I, at least, see the first one much more often. – Akiva Weinberger Mar 18 '15 at 01:32
  • People who don't know what they are doing often write $\sqrt{-1} = i$; these people are mistaken. – MJD Mar 18 '15 at 01:33
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    No, they're just using a branch cut. Just like we do with logs. – Akiva Weinberger Mar 18 '15 at 01:37
  • Note that this disagreement has nothing to do with the OP, as the computation fails in either convention. That said, I'd be interested to see a serious defense by MJD of your most recent comment. – Kevin Carlson Mar 18 '15 at 02:16
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The $\sqrt{\cdot}$ function is problematic in the complex numbers. By definition, for a non-negative real number $r$, $\sqrt{r}$ is the non-negative root, but "negative" is not well-defined for complex numbers.

Your rewriting $\sqrt{-1}$ as $\pm i$ is actually about as sensible a result as you could hope for: it gives you a list including every number whose square is $-1$. A safer way to do the same calculation (avoiding the problematic square root) might be

$$z = i^{-1}$$ $$z = \frac1i$$ $$z^2 = \left(\frac1i\right)^{2}$$ $$z^2 = \frac{1}{-1}$$ $$z^2 = -1$$ $$z = \pm i$$

Instead of slapping a $\sqrt{\cdot}$ sign on a number whose square root isn't defined, these equations use $z^2$ to keep track of the fact that the number we are looking for is a number whose square is the stated number.

If you didn't already know that $-i$ is $i^{-1}$, this would give a good hint that it might be. To really find out, however, you have to try multiplying by $i$ as shown in the answer by ncmathsadist.

David K
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