Rewriting imaginary number expression

**live_laugh_luv27** · January 18th 2009, 12:51 PM

Write the expression (square root of 12) * (square root of -4) / (square root of 3) in the form a+bi, where a and b are real numbers.
Thanks for any help!

**Mush** · January 18th 2009, 12:55 PM

Originally Posted by live_laugh_luv27

Write the expression (square root of 12) * (square root of -4) / (square root of 3) in the form a+bi, where a and b are real numbers.
Thanks for any help!

$\frac{\sqrt{12}\times \sqrt{-4}}{\sqrt{3}}$

$= \frac{\sqrt{12}\times \sqrt{-1 \times 4}}{\sqrt{3}}$

$= \frac{\sqrt{12} \times \sqrt{-1} \times \sqrt{4}}{\sqrt{3}}$

$= \frac{\sqrt{12} \times i \times 2}{\sqrt{3}}$

$= 2i \times \frac{\sqrt{12}}{\sqrt{3}}$

$= 2i \times \sqrt{\frac{12}{3}}$

$= 2i \times \sqrt{4}$

$= 2i \times 2$

$= 4i$

$= 0+4i$

**live_laugh_luv27** · January 18th 2009, 01:07 PM

Thanks!
What if 12 was negative, how would the problem change??
Thanks again for your help.

**Mush** · January 18th 2009, 01:31 PM

Originally Posted by live_laugh_luv27

Thanks!
What if 12 was negative, how would the problem change??
Thanks again for your help.

If 12 was negative then you would do the exact same thing as I did with the -4.

$\sqrt{-12} = \sqrt{12 \times -1}$

$= \sqrt{12} \times \sqrt{-1}$

$= \sqrt{12} \times i$

$= i\sqrt{12}$

In general:

$\sqrt{-a} = i\sqrt{a}$ .

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