Prof, are you saying that in the context of complex numbers that \(\displaystyle \sqrt{25}\) =\(\displaystyle \pm5\) as well?Simple enough: [MATH]\sqrt{-4} = \pm2i[/MATH]. Both answers are correct -- or neither is, since the square root, in the context of complex numbers, is not a single-valued function.
Or, if you choose to define [MATH]\sqrt{-1} = i[/MATH] only, as many algebra books do when they introduce imaginary numbers, then you forfeit the rule that [MATH]\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}[/MATH] for all real numbers a, b. Then the answer has to be [MATH]\sqrt{-4} = 2i[/MATH], and not [MATH]\frac{\sqrt{4}}{\sqrt{-1}} = \frac{2}{i} = -2i[/MATH].
Of course I do not speak for Prof. Peterson, but \(\displaystyle \sqrt{25}=5\) is one number. Now twenty has two square roots they are \(\displaystyle \pm\sqrt{25}=\pm5\) For this prof in his complex variables classes this is what we use: "we add(enlargement of the reals) one number \(\displaystyle \mathit{i}\) that a solution for the equation \(\displaystyle x^2+1=0\). Notice that means \(\displaystyle -\mathit{i}\) is also a solution.Prof, are you saying that in the context of complex numbers that \(\displaystyle \sqrt{25}\) =\(\displaystyle \pm5\) as well?