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# Complex numbers

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273
5 Thank you

Nov 27, 2018

#1
+2

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 !!   Nov 27, 2018
edited by CPhill  Nov 27, 2018
#2
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Would you mind explaining why, as this is exactly what i needed help with. thanks

YEEEEEET  Nov 27, 2018
#3
0

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 ???   CPhill  Nov 27, 2018
#4
+1

yes that makes perfect sense but does that not also apply to 8-2p=0?

as 2pi^2=-2p thank you for responding

YEEEEEET  Nov 27, 2018
#5
+1

Ah....I made a mistake....let me do a corrected post...thanks for pointing out my error !!   CPhill  Nov 27, 2018