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I'm reading Linear Algebra Done Right by Sheldon Axler. In Chapter $1$, Exercise $A$, #2, it states:

Show that $(-1 + \sqrt{3i})/2$ is a cube root of $1$.

The solution on linearalgebras.com shows the following solution here, at number 2.

It states that $(-1 + \sqrt{3i})/2$ squared is $(-1 - \sqrt{3i})/2$. I can understand the rest of the solution but I don't know how they got past the first step. I even squared the above value in an on-line calculator and it gave me a different value. Where am I going wrong?

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    $\begingroup$ The symbol $\sqrt{~}$ is meaningless for a complex number: it denotes the positive square root of a positive real number. $\endgroup$ – Bernard Aug 17 '18 at 8:14
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    $\begingroup$ Most certainly it is $\sqrt{3}\,i$, rather than $\sqrt{3i}$. $\endgroup$ – egreg Aug 17 '18 at 8:54
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The $i$ should not be inside the square root.

It should be $$\left(\frac{-1+\sqrt3 \,i}{2}\right)^{\!2}=\frac{-1-\sqrt3 \,i}{2}$$

rather than

$$\left(\frac{-1+\sqrt{3 i}}{2}\right)^{\!2}=\frac{-1-\sqrt{3 i}}{2}$$

It might be easier to understand it as $$\exp \left(\frac{2\pi i}3 \right)^{\!2}=\exp \left(\frac{4\pi i}3 \right)$$

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  • $\begingroup$ I often type \; just after a square root to get a little space, to make sure that what comes after is clearly seen to be NOT under the square-root symbol......+1 $\endgroup$ – DanielWainfleet Aug 17 '18 at 8:35
  • $\begingroup$ ah, good suggestion. $\endgroup$ – Siong Thye Goh Aug 17 '18 at 8:37
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    $\begingroup$ Better \, than \;. It's a general rule I apply in my TeX code: when a variable or a parenthesis follows a radical, a thin space \, is most likely needed. Also \right)^{\!2} would push the exponent towards the parenthesis. $\endgroup$ – egreg Aug 17 '18 at 8:56
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Consider equation:

$x^3-1=0$

$$x^3-1=(x-1)(x^2+x+1)=0$$

$$x^2+x+1=0$$

$$x=\frac{-1±\sqrt {1-4}}{2}=\frac{-1±\sqrt 3 i}{2}$$

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