-2
$\begingroup$

This question already has an answer here:

Please tell what is wrong with following :

$\iota=\sqrt{-1}$

multiply both sides with $\iota$

$\iota\cdot \iota=\sqrt{-1}\cdot\sqrt{-1}\Rightarrow\iota^{2}=\sqrt{(-1)\cdot(-1)}$

$\Rightarrow \iota^{2}=1$

which is ofcourse a contradiction. I know step

$\sqrt{-1}\cdot \sqrt{-1}=\sqrt{(-1)\cdot(-1)}$

is wrong but how to explain this to students who have just started learning complex numbers??

$\endgroup$

marked as duplicate by Moya, YoTengoUnLCD, Zev Chonoles, carmichael561, Théophile May 11 '16 at 4:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1
$\begingroup$

The problem is that the symbol "$\sqrt{\cdot}$" does not have the clear-cut meaning with complex numbers that it has with positive real numbers. For positive reals, you always select the positive root. No such option exists in general with complex numbers, so the familiar rule $$\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}$$ needn't hold.

$\endgroup$

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