+129852 8 + pi [ 2i + 1 ] 16i + 2pi^2 + 8 + pi
_____ ________ = ________________ =
2i - 1 [2i + 1 ] 4i^2 - 1
8 + (16 + p - 2p) i 8 + ( 16 - p) i
_______________ = ______________
-5 -5
Note YEEEEEET.....that this will only be a real number when 16 - p = 0
So....16 - p = 0...so p = 16
So we will have
8 + 0i 8
____ = - ___
-5 5
CORRECTED ANSWER.....thanks YEEEEEET for spotting my earlier error !!
+845 Would you mind explaining why, as this is exactly what i needed help with. thanks
+129852 We first want to make the denominator a real number
We can do this by multiplying the the numerator/denominator by the conjugate of 2i - 1
This is 2i + 1
Next...we see that to make the numerator a real number....we need to have (16 + p) i =
0i
I see that I can do this if p = - 16
Does that make sense ???
