Trigonometric Methods - Calculating impedance in rectangular and polar forms

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rikiki
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Homework Statement


Given the equivalent impedance of a circuit can be calculated by the
expression

Z= (Z_1 Z_2)/(Z_1+ Z_2 )

If Z1 = 4 + j10 and Z2 = 12 – j3, calculate the impedance Z in
both rectangular and polar forms.

Homework Equations



j2=-1

The Attempt at a Solution



Z= ((4+j10)×(12-j3))/((4+j10)+(12-j3))

Z=(48+j12-j120-j^2 30 )/(16-j7)

Z= (48+j12-j120-(-1)30)/(16-j7)

Z= (48-j108+30)/(16-j7)

Z= (78-j108)/(16-j7)

I've got completely stuck I'm afraid. From searching about, it appears it would be easier to convert this straight into polar form. My notes are quite shocking on this, please see attached. I'm using microsoft mathematic software as my calculator, but having absolutely no joy figuring out how to do this. If anybody can offer some assistance, even in just how to get the answer in the attached I'm sure i can work the rest out. Thanks.
 

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Are you asking for help with:
1. Interpreting your calculator's user instructions.
2. Converting from rectangular to polar form "by hand".
3. Reducing the Z in your last line to the form a + bj.

By the way, check the sign of the imaginary term in the denominator.
 
ok thanks I've gone back to the drawing board and come up with this so far. got too engrossed in trying to find the quick way with a calculator. Hopefully this looks a bit more promising.

Z= (Z_1 Z_2)/(Z_1+ Z_2 )

r= √(a^2+ b^2 )
θ=arctan b/a

Z_1=4+j10
r = √(4^2+ 〖10〗^2 )
r=10.8
θ=arctan 10/4
θ=68.2°

Z_2=12-j3
r= √(〖12〗^2+ 3^2 )
r=12.4
θ=arctan (-3)/12
θ=-14.0°


Z= (Z_1 Z_2)/(Z_1+ Z_2 )

Z= (10.8 ∠68.2 ×12.4 ∠-14.0)/((4+j10)+(12-j3))

Z=(10.8 ×12.4 ∠ 68.2-14.0)/(16-j7)

Z= (133.92 ∠54.2)/(16-j7)

Z=((133.92 ∠54.2)/(17.5 ∠ 23.6))

Z=(133.92)/(17.5) ∠54.2-23.6

Z=7.7 ∠30.6°

just need to find the way to convert to rectangular form now.
 
Watch the signs when you add. In the denominator, j10 - j3 is not -j7. I see that you managed to "correct" the error when you found the angle for the denominator.