How does 12 + j16 become 20 ∠53.1° in impedance calculations?

[portcharlotte]

Good morning,

I was reviewing a problem in my book regarding Impedance. I have a question. For the impedance -Z they got 20 ∠53.1° degrees How did they go from 12 + j16 to 20 ∠53.1°. Thanks

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[I like Serena]

Good morning,

I was reviewing a problem in my book regarding Impedance. I have a question. For the impedance -Z they got 20 ∠53.1° degrees How did they go from 12 + j16 to 20 ∠53.1°. Thanks

Hi portcharlotte, welcome to MHB! ;)

Every imaginary number has a cartesian representation and a polar representation.
The relation between $x+jy$ and $r∠\theta$ is given by $r=\sqrt{x^2+y^2}$ and $\theta=\arctan\frac yx$.
That is, unless $x\le 0$ in which case we have to select the appropriate $\theta$ in the unit circle, as it is out of range of $\arctan$.

In this case we have $x=12$ and $y=16$. Therefore $r=\sqrt{12^2+16^2}=20$ and $\theta=\arctan\frac{16}{12}\approx 53.1°$

 

[portcharlotte]

Thank you so much Professor.