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Consider I am having a complex number:

$x = 4i$.

  1. Squaring on both sides: $x^2 = -16$.
  2. Multiply by $-1$ on both sides: $-x^2 = 16$.
  3. Taking square root on both sides: $xi = 4$.

What I am doing wrong here?
Why $i$ is not in denominator?

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    $\begingroup$ That should rather be $xi=\pm 4$ at the last step. $\endgroup$ – dxiv Mar 17 '17 at 17:55
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    $\begingroup$ Actually, applying the quadratic formula, we find that:$$-x^2=16\implies x=\frac{\pm\sqrt{-64}}{-2}=\pm4i$$By checking extraneous solutions, $x=-4i$ is incorrect and $x=4i$ is correct. $\endgroup$ – Simply Beautiful Art Mar 17 '17 at 17:56
  • $\begingroup$ True, but i is not in denominator that is my biggest concern. $\endgroup$ – MinusInfinity Mar 17 '17 at 17:56
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    $\begingroup$ @hariharan Note: $-i=\dfrac1i $ $\endgroup$ – Mark S. Mar 17 '17 at 18:27
  • $\begingroup$ @MarkS. I guess that answers my question $\endgroup$ – MinusInfinity Mar 17 '17 at 18:36
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The last step should be $x_i= \pm 4$. Because the square root of a number is either its positive or negative value. Recognize that

$4 \times 4 = 16$

and

$-4 \times -4 = 16$.

However, now you can check your answer with the first line $x=4i$. $x \times i = 4i \times i = -4$. Thus $xi=-4$.

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    $\begingroup$ $x = -4$ is clearly not a solution. $\endgroup$ – Qudit Mar 17 '17 at 18:17
  • $\begingroup$ Sorry it was $xi = -4$ $\endgroup$ – JahKnows Mar 17 '17 at 18:25
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In the reals, $\sqrt{x^2}=x$ doesn't hold.

Similarly, there is no reason that in the complex $\sqrt{-x^2}=ix$ holds.

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  1. Squaring on both sides: $x^2 = -16$.

So far so good.

  1. Multiply by $-1$ on both sides: $-x^2 = 16$.

Here you need to exercise some care. $(-1)(x^2) = -(x^2)$. Then the next step needs to be amended:

  1. Taking square root on both sides: $\sqrt{-(x^2)} = 4$.

Okay, wait, where exactly are you going with this? You need to be clear on the difference between the principal square root and the other square root. If you start with $x = -4i$, you might end up with the same confusion.

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