Let $z_1 = r_1 \exp(i \theta_1) $ and $z_2 = r_2\exp(i \theta_2) $ be complex numbers in exponential form and $z = z_1 + z_2 $, then I know using the complex plane vector representation that $z$ can be written in the exponential form too, $z = r \exp(i \theta) $, where $$ r = \sqrt{r_1^2 + r_2^2 + 2 r_1 r_2 \cos( \theta_1 - \theta_2)} $$ $$ \theta = \arctan\left(\frac{r_1 sin\theta_1 + r_2 sin\theta_2} {r_1 cos\theta_1 + r_2 cos\theta_2}\right) $$

Problem: when I plotted-in Mathematica- the real part of $z$, it did not match with the plot of the real part of $z_1 + z_2$ but rather it matched the plot of the real part of its absolute value $|{z_1 + z_2}|$. I tried two different numbers for $z_1, z_2$, both have the problem.

Question: 1) Why is that, what is wrong with the formula?, 2) How to fix it so both $Re[z_1 + z_2]$ and $Re[z]$ give the same plot?

  • 1
    $\begingroup$ I don't know if this will solve your problem, but try this: the arctangent formula for the angle only works when the angle of the resulting complex number is between $-\pi/2$ and $\pi/2$. Try adjusting the whole thing by an angle of $\pi$. $\endgroup$
    – Josh B.
    Jul 20, 2020 at 19:09

1 Answer 1


$z = z_1+z_2 = r_1\cos\theta_1 + r_2\cos\theta_2 + ir_1\sin\theta_1+ir_2\sin\theta_2$.

$\mod(z) = \sqrt{(Re(z))^2 + (Im(z))^2} = \sqrt{r_1^2(\cos^2\theta_1 + \sin^2\theta_1) + r_2^2(\cos^2\theta_2 + \sin^2\theta_2) + 2r_1r_2(\cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2)} = \sqrt{r_1^2+r_2^2+2r_1r_2\cos(\theta_1-\theta_2)}$.

$\arg(z) = \arctan(Im(z)/Re(z)) = \arctan(\frac{r_1\sin\theta_1 + r_2\sin\theta_2}{r_1\cos\theta_1 + r_2\cos\theta_2})$.

Looks correct to me. Maybe you are plugging into Mathematica incorrectly? Or possibly the arctan function in Mathematica is defined differently?


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