simplify:

3/i + 6/(i+2) + i/(i-2)

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- Nov 7th 2006, 04:49 PMMr_Greensimplify:
simplify:

3/i + 6/(i+2) + i/(i-2) - Nov 7th 2006, 04:56 PMtopsquark
i is a square root. Thus we wish to get rid of the square roots in the denominators. You do this by multiplying the denominator by the complex conjugate of the denominator.

For example:

$\displaystyle \frac{6}{i+2} = \frac{6}{i+2} \cdot \frac{-i + 2}{-i + 2}$

= $\displaystyle \frac{6(-i+2)}{-i^2 + 4} = \frac{12 - 6i}{1+4} = \frac{12-6i}{5}$

Can you take it from here?

-Dan - Nov 7th 2006, 07:11 PMMr_Green
- Nov 7th 2006, 08:14 PMJameson
Well he did one of the terms. Are you paying attention to his method? The method is the key.

$\displaystyle \frac{3}{i} \times \frac{i}{i}=-3i$

$\displaystyle \frac{i}{i-2} \times \frac{i+2}{i+2}=...$