I am just learning more about complex numbers and a question popped up I can't figure out on my own, so I've posted it here. I already know $i^2=-1$ and $i=\sqrt{-1}$ (isn't it even true that $\pm i=\sqrt{-1}$?)

I know $\sqrt{a} \sqrt{b} = \sqrt{ab}$ is only defined or valid if at least one of the two is a positive number. One of the both being negative should work as well, see

(1) $\sqrt{3} \sqrt{2} = \sqrt{6}$

(2) $\sqrt{3} \sqrt{-2} = \sqrt{-6} = \sqrt{(-1)6}=\pm\sqrt{6}i$

Now the tricky part:

What about

(3) $\sqrt{-3i} = \sqrt{(-1)3i}$

(4) $\sqrt{(-1)3i}=\sqrt{-1}\sqrt{3i}=i\sqrt{3i}$

The third (3) is true (obviously), but Wolfram Alpha says the fourth (4) is not true anymore. Can anyone tell me why? Assuming $a=-1$ (negative) and $b=3i$ (positive) the formula above should be working, or am I wrong?

Best regards!

  • $\begingroup$ Neither $-1$ nor $3i$ is a positive number, so there is no expectation that the square root should distribute. There is no notion for (non-real) complex numbers to be "positive" or "negative": indeed whenever someone writes "$z>0$" it's tacitly implied that $z$ is a real number. $\endgroup$ – Erick Wong Jul 24 '15 at 13:34
  • $\begingroup$ Hey @ErickWong, thank you! I did not know that complex numbers (or better: a complex number with an imaginary part greater than 0) can not be ordered into "negative" and "positive" numbers. Now I know :) $\endgroup$ – David Becher Jul 24 '15 at 13:44

First, how would you define positive and negative for complex numbers? For example, is $-1+2i$ positive of negative?

Second, when dealing with complex numbers, we often define the square root to be multivalued. $$\sqrt{-3i}=\pm \sqrt{3} i \sqrt{i}$$ is true.

Please check This and This.

  • $\begingroup$ Hey @wythagoras I typed in your equation into Wolfram Alpha and it says, this equates to false, see Link 1 However, this seems to be true instead Link 2 $\endgroup$ – David Becher Jul 24 '15 at 13:47
  • $\begingroup$ @DavidBecher W|A inteprented it as something else. Please check the links added to the answer. $\endgroup$ – wythagoras Jul 24 '15 at 13:53
  • $\begingroup$ @DavidBecher Notice the $\pm$ I wrote. W|A used the principle root, which has a negative sign. The positive sign is another possibility. You can find it all in the links I posted. $\endgroup$ – wythagoras Jul 24 '15 at 13:54

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