(Crying)how do you write 68/4+i - 17/4-i in a+bi format?

can anyone explain not just answer this ?

pls :)

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- Nov 21st 2008, 05:50 AMIronscorppre- calc test next period need help please ...
*(Crying)how do you write 68/4+i - 17/4-i in a+bi format?*

can anyone explain not just answer this ?

pls :) - Nov 21st 2008, 09:54 AMTwighi
I assume you mean 68/4+i to mean: $\displaystyle \frac{68}{4+i}$

We get:

$\displaystyle \frac{68}{4+i} - \frac{17}{4-i}$

Then you do something called multiply with the complex conjugat.

You multiply both numerator and denominator with (focusing on the first term), you multiply with $\displaystyle (4-i)$

So we get(still focusing on the first term):

$\displaystyle \frac{68(4-i)}{(4+i)(4-i)} $

Simplify this... $\displaystyle \frac{68(4-i)}{17} $

You do the same thing on the second term, and add them togheter.

Hope this helped you out - Nov 25th 2008, 05:28 AMIronscorp
yeah thanks :)