I was looking at a problem $\sqrt{x}=-3$, and I had at first thought $x=9 i^4$ was a solution. ($\sqrt{9 i^4}=3i^2=-3$)

Though I then realized that this would cause some problems.

For example using this, we would have $\sqrt{i^4}=i^2=-1$. While on the other hand $\sqrt{i^4}=\sqrt{1}=1$.

I checked Wolfram and it says that $\sqrt{i^4} \neq i^2$ (also $(i^4)^{1/2}$). Could any one explain to me why we can't treat the exponents of $i$ this way?

Is it possible to algebraically show that $\sqrt{x}=-3$ has no solutions?

I am trying to learn some complex analysis and this made me realize that I might have some really bad intuition on complex numbers.

  • $\begingroup$ sqrt isn't a defined function on the complex place $\endgroup$ May 9 '17 at 4:40
  • 2
    $\begingroup$ The rule $(z^a)^b=z^{ab}$ does not always hold on the complex plane if $a$ and $b$ are not both integral. $\endgroup$ May 9 '17 at 4:43
  • $\begingroup$ $\sqrt{x}=-3$ can have solutions but they depend on your definition of the square-root in the complex plain. In particular, you have to define a branch cut from $z=0$ to $z=\infty$ in order to render $\sqrt\cdot$ a complex function. $\endgroup$
    – Fabian
    May 9 '17 at 4:46
  • $\begingroup$ This is almost the same as math.stackexchange.com/questions/49169/… . The answers there may help. $\endgroup$ May 9 '17 at 4:46
  • $\begingroup$ Thank you very much for the replies. $\endgroup$ May 9 '17 at 4:48

It's the same reason as why $\sqrt{(-1)^2}\neq -1$, no need to involve complex numbers in this case (to complicate our lives). When we consider square root as function $\sqrt{\,\cdot\,}\colon\mathbb R_{\geq 0}\to \mathbb R_{\geq 0}$, we define it as a function with property $(\sqrt x)^2=x$. On the other hand, $\sqrt{x^2}\neq x$ in general, it fails whenever $x<0$. We actually have $\sqrt{x^2} = |x|$.


The square root function is defined as being positive. As you likely know, $i^2$ is, by definition, negative.

Given the above, $\sqrt{x}=-3$ would have no solutions.

This is different from solving for $x^2 = 9$. There are two solutions,one of which is $-3$.

  • $\begingroup$ If I had written it as $x^{1/2}=-3$, would it be any different? $\endgroup$ May 9 '17 at 4:53

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