For the following three phasors


A = 6 – j8
B = 60 + j80
C = 30 – j4


Perform the flowing operations and give the final answers in Polar form:

A + B – C

Attempt:

A + B - C
(6 - j8) + (60 + j80) - (30 - j40)
36 - j8 + j80 + j40

M = (A^2 + B^2)^0.5
=((6 - j8)^2 + (60 + j80)^2)^0.5

I carried on simplifying this then got the following:

(3636 + 9504j + 6464j^2)^0.5

I am not sure what to do next

 

Not quite
The real part of A + B - C is (6 + 60 - 30) = 36
The imaginary part of A + B - C is (-8 + 80 - (-4)) = 76
So the resultant phasor is 36 + j76

|A+B-C| = √(36)^2 + (76)^2) = √(1296 + 5776) = √(7072) ≈ 84.1

The resultant phasor is in the first quadrant so the angle will be
> 0° and < 90°

The angle is arctan(76/36) ≈ 64.65°

So in polar form it would be:

\(84.1\angle64.65^{\circ}\)

If C is actually 30 - j40 you should be able to follow my example for 30 - j4 above