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We know that $\sqrt {-1}=\{i,-i\}$, then $\sqrt {-1} \sqrt {-1}=\{i,-i\}\{i,-i\}=\{1,-1\}$ or $\sqrt {-1} \sqrt {-1}=\{i^2, (-i)^2\}=\{-1\}$?

Omar Dennaoui
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Arvind
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2 Answers2

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One way make sure no funny business is going on let's use Euler's formula: $$e^{i \pi } +1 =0$$ so $-1 = e^{i \pi }$, but remember that's not the only way to express $-1$. Indeed $\pi$ radians around the unit circle is the same as $3 \pi$ or $5 \pi$ radians etc $\cdots$. So the correct way to express $-1$ is: $$-1 = e^{i(2k+1)}$$ where $k \in \mathbb Z$.

Now that's covered, let's look at \begin{align} \sqrt {-1} \sqrt {-1} &= \sqrt {e^{i(2k+1)}} \sqrt {e^{i(2k+1)}} \\ &={e^{i(2k+1)}}^{\frac 12} {e^{i(2k+1)}}^{\frac 12} \\ &=e^{i\frac{2k+1}{2}} e^{i\frac{2k+1}{2}} \ \ , \text {using properties of exponents} \\ &=e^{i(\frac{2k+1}{2} + \frac{2k+1}{2})}, \text {using properties of $e$} \\ &=e^{i(2k+1)} \\ &= -1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ ,\text {by definition} \end{align}

Omar Dennaoui
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  • the best advice one can give regarding $\mathbb C$ when there are multiple values to look out for is to right the general form of your expression, i.e. $-1 = e^{i (2k+1)\pi}$ – Omar Dennaoui May 24 '22 at 22:49
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Following Omar's advice, $\sqrt{-1}=e^{i(2k+1)\pi/2}$ for $k\in\mathbb{Z}$, we get

$\sqrt{-1}\sqrt{-1}=e^{i(2m+1)\pi/2} e^{i(2n+1)\pi/2}$,

and not $= e^{i(2k+1)\pi/2} e^{i(2k+1)\pi/2}$.

And that simplifies to $e^{ik\pi}=\{1,-1\}$.

This has connection to the anomally that

in general, $\log u+\log u\neq 2\log u$ for a complex numbers $u$.

Arvind
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