How do I evaluate complex numbers in rectangular form and polar form?

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SUMMARY

The discussion focuses on evaluating complex numbers in both rectangular and polar forms, specifically the expression \(\frac{10 + j5 + 3\angle 40^{\circ}}{-3 + j4} + 10 \angle 30^{\circ}\). The correct solution is established as \(8.293 + j2.2\), although one participant initially arrived at a different answer due to errors in converting to polar form. Key insights include the importance of maintaining the correct signs when applying the arctangent function during conversion and the recommendation to keep the denominator in rectangular form for simplification.

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Homework Statement


Evaluate the following complex numbers (results in rectangular form):
\frac{10 + j5 + 3\angle 40^{\circ}}{-3 + j4} + 10 \angle 30^{\circ}

Homework Equations


I know that the solution is:
8.293 + j2.2
But i get a different answer.

The Attempt at a Solution


Convert into rectangular form when adding or subtracting and to polar form when multiplying and dividing. Change the answer to rectangular form.
\frac{10 + j5 + 3\angle 40^{\circ}}{-3 + j4} + 10 \angle 30^{\circ}

\frac{10 + j5 + 3cos(40^{\circ}) + 3jsin(40^{\circ})}{[(-3)^2 + (4)^2]^1^/^2 \angle arctan(4/-3)} + 10cos(30^{\circ}) + 10jsin(30^{\circ})

\frac{10 + j5 + 2.30 + j1.93}{5 \angle -53.13^{\circ}} + 8.66 + j5

\frac{12.30 + j6.93}{5 \angle -53.13^{\circ}} + 8.66 + j5

\frac{[(12.30)^2 + (6.93)^2]^1^/^2 \angle arctan(6.93/12.30)}{5 \angle -53.13^{\circ}} + 8.66 + j5

\frac{14.12 \angle 29.40^{\circ}}{5 \angle -53.13^{\circ}} + 8.66 + j5

\frac{14.12}{5}\angle [29.40^{\circ}-(-53.13^{\circ})] + 8.66 + j5

2.824 \angle 83.53^{\circ} + 8.66 + j5

0.37 + j2.8 + 8.66 + j5

9.03 + j7.8
 
Last edited:
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Why did you convert the denominator to polar form? Keep it in rectangular form, multiply numerator and denominator by the complex conjugate of the denominator to rationalize the fraction, and go from there...
 
I converted it all in polar form because i believed that no matter if i use it in polar or rectangular form, the answer must be the same.

I found out that my error was when i was converting it to polar form. I forgot that after applying the arctangent to verify that the phase of the angle would give the correct signs otherwise i would have to shift the phases by 180 degrees.
 

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