+0  
 
0
273
5
avatar+845 

Thank you

 Nov 27, 2018
 #1
avatar+107405 
+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 !!

 

cool cool cool

 Nov 27, 2018
edited by CPhill  Nov 27, 2018
 #2
avatar+845 
0

Would you mind explaining why, as this is exactly what i needed help with. thanks

YEEEEEET  Nov 27, 2018
 #3
avatar+107405 
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 ???

 

 

cool cool cool

CPhill  Nov 27, 2018
 #4
avatar+845 
+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
avatar+107405 
+1

Ah....I made a mistake....let me do a corrected post...thanks for pointing out my error !!

 

cool cool cool

CPhill  Nov 27, 2018

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