0
$\begingroup$

Correct me if I am wrong, $\sqrt{-4}=2i$.

But how do you explain it to a student?

We know $\sqrt{-1}=i$, but one cannot say $\sqrt{-4}=\sqrt{-1}\sqrt{4}=2i$ as the laws of indices can only be applied to real numbers.

Any advice would help.

$\endgroup$
  • 4
    $\begingroup$ Write it as $-4=(2i)^2$. $\endgroup$ – Lord Shark the Unknown May 14 at 5:09
  • $\begingroup$ How would you answer a student if he ask why choose the positive root and not the negative? $\endgroup$ – LanaDR May 14 at 5:11
  • 4
    $\begingroup$ Neither $2i$ nor $-2i$ are positive. $\endgroup$ – Lord Shark the Unknown May 14 at 5:12
  • $\begingroup$ that is what I thought so as positive or negative does not make sense for complex numbers $\endgroup$ – LanaDR May 14 at 5:13
  • 1
    $\begingroup$ What is $\sqrt{-4}$? Whatever it is, it is not "positive". $\endgroup$ – Lord Shark the Unknown May 14 at 5:15
2
$\begingroup$

Edit:

$i^2 = -1$

Therefore,

$4i^2 = -4$

and by taking square root on both sides, we have $2i = \sqrt{-4}$

$\endgroup$
  • $\begingroup$ sounds good! thanks! $\endgroup$ – LanaDR May 14 at 5:18
  • 2
    $\begingroup$ How is this any different from $\sqrt{-4}=\sqrt{-1}\sqrt{4}$, which was rejected in the question? $\endgroup$ – amd May 14 at 5:24
  • $\begingroup$ Even if it isn't, why is it wrong? $\endgroup$ – NoLand'sMan May 14 at 5:28
  • 2
    $\begingroup$ One needs to be careful to know what is defined for the complex numbers. For example, $\sqrt{ab} = \sqrt a\sqrt b$ is not necessarily true for complex numbers. $\endgroup$ – user1952500 May 14 at 5:31
  • $\begingroup$ Okay, I have edited my answer, is this any better? $\endgroup$ – NoLand'sMan May 14 at 5:37
1
$\begingroup$

The square root of a negative number is usually given by $$\begin{align} (-4)^{1/2} &=e^{\ln{(-4)}/2}\\ &=e^{\ln{(4e^{i\pi})}/2}\\ &=e^{\ln{(e^{i\pi+2\ln{(2)}})}/2}\\ &=e^{\ln{(2)}+i\pi/2}\\ &=2e^{i\pi/2}\\ &=2\left(\cos{\left(\frac\pi2\right)}+i\sin{\left(\frac\pi2\right)}\right)\\ &=2i\\ \end{align}$$ where the principal branch of the natural logarithm is taken instead of the 'positive' square root.

$\endgroup$
0
$\begingroup$

I would approach the topic dealing first with the $2 \pi$ periodicity of the angle.

Then I would pass to the convention used to split the $2 \pi$ angle: the "standard" is $-\pi < \alpha \le \pi$ thus the "principal" "standard" solution is $2i$.
But any other split can be adopted.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.