• August 15th 2012, 11:29 PM
rich1979
Y=1/Z1+1/Z2+1/Z3

Z1= 2+j2,
Z2=1+j5 ,
Z3=j6

i cant figure out how to find Y, could someone please explain how to,or just point me in the right direction. I would be very grateful for any help on this??Thanks!!
• August 15th 2012, 11:51 PM
MaxJasper
Y= (1-j)/2
• August 16th 2012, 12:14 AM
rich1979
• August 16th 2012, 08:25 AM
Soroban
Hello, rich1979!

Quote:

$\text{Given: }\:\begin{Bmatrix}z_1 &=& 2+2i \\ z_2 &=& 1+5i \\ z_3 &=& 6i \end{Bmatrix}$

$\text{Find: }\:y \:=\:\frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3}$

We must rationalize those denominators.

$\frac{1}{z_1} \;=\;\frac{1}{2+2i} \;=\;\frac{1}{2+2i}\cdot\frac{2-2i}{2-2i} \;=\;\frac{2-2i}{8} \:=\:\frac{1}{4} - \frac{1}{4}i$

$\frac{1}{z_2} \;=\;\frac{1}{1+5i} \;=\;\frac{1}{1+5i}\cdot\frac{1-5i}{1-5i} \;=\;\frac{1-5i}{26} \;=\;\frac{1}{26} - \frac{5}{26}i$

$\frac{1}{z_3} \;=\;\frac{1}{6i} \;=\;\frac{1}{6i}\cdot\frac{-6i}{-6i} \;=\;\frac{-6i}{36} \;=\;-\frac{1}{6}i$

$\text{Hence: }\:y \;=\;\left(\frac{1}{4}-\frac{1}{4}i\right) + \left(\frac{1}{26} - \frac{5}{26}y\right) + \left(-\frac{1}{6}i\right)$

n . . . . $y \;=\;\left(\frac{1}{4} + \frac{1}{26}\right) - \left(\frac{1}{4} + \frac{5}{26} + \frac{1}{6}\right)i$

n . . . . $y \;=\;\frac{15}{52} - \frac{95}{156}i$
• August 16th 2012, 09:15 AM
MaxJasper
I assumed j2, j5, j6 mean j^2, j^5, j^6
• August 16th 2012, 04:25 PM
rich1979
• August 16th 2012, 04:28 PM
rich1979
maxjasper no j2 is a complex number.
• August 16th 2012, 04:46 PM
MaxJasper
Quote:

Originally Posted by rich1979
maxjasper no j2 is a complex number.

Ok, that means j2=a+b*i where i=sqrt(-1). We can write j5, j6 in similar manner and then simplify y:
$\text{j2}\text{:=}a+b I;\text{j5}\text{:=}c+d I;\text{j6}\text{:=}e+f I;\text{z1}\text{:=}2+\text{j2};\text{z2}\text{:=} 1+\text{j5};\text{z3}\text{:=}\text{j6};y\text{:=} \frac{1}{\text{z1}}+\frac{1}{\text{z2}}+\frac{1}{\ text{z3}}$

Then we will have:

$y=\frac{2}{(2+a)^2+b^2}+\frac{a}{(2+a)^2+b^2}+\fra c{1}{(1+c)^2+d^2}+\frac{c}{(1+c)^2+d^2}+\frac{e}{e ^2+f^2}+i \left(-\frac{b}{(2+a)^2+b^2}-\frac{d}{(1+c)^2+d^2}-\frac{f}{e^2+f^2}\right)$