pre- calc test next period need help please ...

Ironscorp · #1 $Old$ November 21st, 2008, 07:50 AM

how do you write 68/4+i - 17/4-i in a+bi format?
can anyone explain not just answer this ?
pls

Twig · #2 $Old$ November 21st, 2008, 11:54 AM

I assume you mean 68/4+i to mean: $\frac{68}{4+i}$

We get:
$\frac{68}{4+i} - \frac{17}{4-i}$
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 $(4-i)$

So we get(still focusing on the first term):
$\frac{68(4-i)}{(4+i)(4-i)}$
Simplify this... $\frac{68(4-i)}{17}$

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

Hope this helped you out

Ironscorp · #3 $Old$ November 25th, 2008, 07:28 AM

yeah thanks