Simple question about imaginary numbers and roots

<p>Okay, if we have square root (a) * square root (b) the answer is equal to square root (ab). And [square root (ab)]^2 = ab. For example: square root (3) * square root (5) = square root (15). This is also true if the numbers are similar: square root (3) * square root (3) = square root (9) = 3</p>

<p>My question is this: why isn't square root (-1) * square root (-1) = square root (-1 * -1) = square root (1) = 1??? I know the answer should be -1 because [square root (-1)] ^2 = -1 because the power cancels the root. But I just want to know why is my way of thinking wrong because I just know it like that "The power cancels the root" but have no real understanding of the matter. Please make your answer as detailed as possible. Thanks in advance!</p>

<p>Simple.
root (a) * square root (b)= square root (ab) doesn’t apply to imaginary numbers.</p>

<p>For Imaginary Numbers
root (a) * square root (b)= -square root (ab)</p>

<p>Thanks for your reply, but you still didn’t help me.</p>

<p>Not that simple (kp2241 isn’t quite right).</p>

<p>If a and b are not negative, then sqrt(ab) = sqrt(a) * sqrt(b).</p>

<p>But if a and/or b can be negative, the rule above may or may not be true.</p>

<p>For example, if sqrt(-4) * sqrt(-9) = 2i * 3i = -6 but sqrt(-4*-9)=6.</p>

<p>On the other hand, sqrt(-4) * sqrt(9) = 2i * 3 = 6i and sqrt(-4*9) = 6i.</p>

<p>Some rules that you have learned for real numbers just don’t work in complex numbers. For example, in the real numbers x^3 = 8 has one solution; in complex numbers it has three solutions.</p>