I was doing numericals on synchronous generators and came across this step in one of the examples. I have no idea what kind of math is used here. Can someone help?

$(1.5 + 2.0j)\Omega = 2.5 \angle 53.13^\circ \Omega$

$(0.3+1.22j)\Omega = 1.256 \angle 76.18^\circ\Omega$

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    $\begingroup$ Your numbers are $a+i b$ and the first number in the rhs is $\sqrt{a^2+b^2}$. Now, guess why and what is the other number ! Happy New Year !! $\endgroup$ Dec 31, 2014 at 14:19
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    $\begingroup$ It looks like complex numbers, only with $j$ instead of $i$. Then on the right side the state the absolute value and the argument of the numbers. $\endgroup$
    – Kaladin
    Dec 31, 2014 at 14:21
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    $\begingroup$ Polar coordinates on the right. $\endgroup$
    – mvw
    Dec 31, 2014 at 14:38
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    $\begingroup$ @Kaladin work with electricity/circuits/etc will use $j$ for the imaginary number because $i$ is used to represent current. $\endgroup$
    – Brian J
    Dec 31, 2014 at 18:06
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    $\begingroup$ @BrianJ Or because engineers have bad handwriting! $\endgroup$
    – rlms
    Dec 31, 2014 at 20:05

1 Answer 1


These are two repesentations of complex numbers (in this case, used to define impedances- see Wikipedia for more info).

Also, note that in electrical engineering and related fields, the imaginary unit is represented as $j$ because the common representation of current is $i$.

Assuming you have a complex number $z$ represented in the cartesian form $a+jb$, you can transform it into other representations, such as polar representation: $$z=a+jb=R\cdot\mathrm{e}^{j\theta}$$ where $$R=|z|=\sqrt{a^2+b^2}\qquad \theta=\arg(z)=\arctan\biggl(\frac{b}{a}\biggr)$$

The representation you show (I do not know the proper name for this representation) is another polar representation, written differently: $$z=a+jb=R\cdot\mathrm{e}^{j\theta}=R\angle\theta$$ where, again, $$R=|z|=\sqrt{a^2+b^2}\qquad \theta=\arg(z)=\arctan\biggl(\frac{b}{a}\biggr)$$

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    $\begingroup$ $\arg z=\arctan(b/a)$ is not really valid for all cases. $\endgroup$
    – egreg
    Dec 31, 2014 at 14:39
  • $\begingroup$ Got it. Thanks :) and Happy New Year to all $\endgroup$ Dec 31, 2014 at 14:47
  • $\begingroup$ @egreg That's right, but as an electrical engineer and former TA of several courses of that carrer, I know that that's generally true in these cases. Also, I prefered to be clear in the scope of the OP's question. $\endgroup$
    – cjferes
    Dec 31, 2014 at 14:53

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