Consider I am having a complex number:
$x = 4i$.
- Squaring on both sides: $x^2 = -16$.
- Multiply by $-1$ on both sides: $-x^2 = 16$.
- Taking square root on both sides: $xi = 4$.
What I am doing wrong here?
Why $i$ is not in denominator?
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5 
- 1$\begingroup$ That should rather be $xi=\pm 4$ at the last step. $\endgroup$dxiv– dxiv2017-03-17 17:55:06 +00:00Commented Mar 17, 2017 at 17:55
- 1$\begingroup$ 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– Simply Beautiful Art2017-03-17 17:56:16 +00:00Commented Mar 17, 2017 at 17:56
- $\begingroup$ True, but i is not in denominator that is my biggest concern. $\endgroup$MinusInfinity– MinusInfinity2017-03-17 17:56:59 +00:00Commented Mar 17, 2017 at 17:56
- 2$\begingroup$ @hariharan Note: $-i=\dfrac1i $ $\endgroup$Mark S.– Mark S.2017-03-17 18:27:48 +00:00Commented Mar 17, 2017 at 18:27
- $\begingroup$ @MarkS. I guess that answers my question $\endgroup$MinusInfinity– MinusInfinity2017-03-17 18:36:40 +00:00Commented Mar 17, 2017 at 18:36
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3 Answers
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2 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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- Squaring on both sides: $x^2 = -16$.
So far so good.
- 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:
- 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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In the reals, $\sqrt{x^2}=x$ doesn't hold.
Similarly, there is no reason that in the complex $\sqrt{-x^2}=ix$ holds.