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

#### LucasE

##### New member
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
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
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
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
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

##### Full Member
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
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
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