Z = (3-i)/(2+pi) and p is real. If Re(Z) = 1/2, find value(s) of p

LucasE

New member
Joined
Feb 15, 2019
Messages
7
Hi guys,
I have the following question:
Z = (3-i)/(2+pi). The value of P must be real. If real value of Z = 1/2. find the value(s) of P.
First I have used the conjugate of (2+pi) in order to rationalize the denominator:
Z = (3-i)(2-pi)/(2+pi)(2-pi)
so: Z = pi^2 -2i - 3pi + 6/ 4-p^2i^2
But, if i^2 = -1, then:
Z = -p -2i -3pi + 6/ 4 + p^2
Where do I go from here?
Thanks
 

topsquark

Full Member
Joined
Aug 27, 2012
Messages
265
Hi guys,
I have the following question:
Z = (3-i)/(2+pi). The value of P must be real. If real value of Z = 1/2. find the value(s) of P.
First I have used the conjugate of (2+pi) in order to rationalize the denominator:
Z = (3-i)(2-pi)/(2+pi)(2-pi)
so: Z = pi^2 -2i - 3pi + 6/ 4-p^2i^2
But, if i^2 = -1, then:
Z = -p -2i -3pi + 6/ 4 + p^2
Where do I go from here?
Thanks
Don't bother with rationalizing the denominator.
\(\displaystyle Z = \dfrac{3 - i}{2 + p i}\)

\(\displaystyle (2 + p i)Z = 3 - i\)

Just solve for p.

-Dan
 

LucasE

New member
Joined
Feb 15, 2019
Messages
7
So,
if i substitute Z for 1/2:
(2+pi)(1/2) = 3-i
1 + 1/2pi = 3 - i
1/2pi = 2 - i
p = 2-i/1/2i
p = (4 - 2i)/i ?
 

Harry_the_cat

Full Member
Joined
Mar 16, 2016
Messages
943
Hi guys,
I have the following question:
Z = (3-i)/(2+pi). The value of P must be real. If real value of Z = 1/2. find the value(s) of P.
First I have used the conjugate of (2+pi) in order to rationalize the denominator:
Z = (3-i)(2-pi)/(2+pi)(2-pi)
so: Z = pi^2 -2i - 3pi + 6/ 4-p^2i^2
But, if i^2 = -1, then:
Z = (-p -2i -3pi + 6)/ (4 + p^2)
Where do I go from here?
Thanks
Ok I've inserted some missing brackets.
Now split your Z up into Real and imaginary parts and let Re(Z) =1/2 and solve for p.
 

Harry_the_cat

Full Member
Joined
Mar 16, 2016
Messages
943
So,
if i substitute Z for 1/2:
(2+pi)(1/2) = 3-i
1 + 1/2pi = 3 - i
1/2pi = 2 - i
p = 2-i/1/2i
p = (4 - 2i)/i ?
No! Z is not 1/2, Re(Z) is 1/2.
 

Harry_the_cat

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Joined
Mar 16, 2016
Messages
943
Don't bother with rationalizing the denominator.
\(\displaystyle Z = \dfrac{3 - i}{2 + p i}\)

\(\displaystyle (2 + p i)Z = 3 - i\)

Just solve for p.

-Dan
I think you do need to rationalise the denominator. Re(Z) =1/2 not Z=1/2.
 

LucasE

New member
Joined
Feb 15, 2019
Messages
7
So, if Z = x + iy,
then if I rearrange (-p-2i-3pi+6):
-p+6 -2i-3pi,
z = (-p + 6 + i(-2-3p))/(4+p^2)
and so, the real part of Z is x:
(-p+6)/(4+p^2)=1/2?
This would suggest that:
-p+6 = 1/2(4+p^2)
so 0 = -4+3/2p^2
where p = 1.633 or -1.633
 

Harry_the_cat

Full Member
Joined
Mar 16, 2016
Messages
943
So, if Z = x + iy,
then if I rearrange (-p-2i-3pi+6):
-p+6 -2i-3pi,
z = (-p + 6 + i(-2-3p))/(4+p^2)
and so, the real part of Z is x:
(-p+6)/(4+p^2)=1/2?
This would suggest that:
-p+6 = 1/2*(4+p^2) … correct
so 0 = -4+3/2p^2 …. not sure how you got this??
where p = 1.633 or -1.633
\(\displaystyle \frac{-p+6}{4+p^2} = \frac{1}{2}\)
\(\displaystyle 2(-p+6)=4+p^2\)
\(\displaystyle -2p+12 = 4 + p^2\)

which is a quadratic equation .... leaving for you to finish
 
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