Modulus of complex number that satisfies $z+\frac{k}{z}=a+bi$

Solution 1:

(Too long for a comment.) $\;$ It is technically possible to derive an equation for $\,|z|\,$ without first solving for $\,z\,$ (as outlined below), but it is more work than solving the quadratic directly.

Let $\,\alpha=a+bi\,$, then after multiplying by $\,z \ne 0\,$ the equation becomes $z^2 - \alpha z + k = 0\,$, and:

• multiplying by $\,\bar z^2 \ne 0\,$: $\;\;|z|^4 - \alpha |z|^2 \bar z + k \bar z^2 = 0$

• taking conjugates and rearranging: $\;\;k z^2 - \bar \alpha |z|^2 z + |z|^4 = 0$

• substituting $\,z^2 = \alpha z - k\,$: $\;\;\left(k \alpha- \bar \alpha |z|^2\right) z + |z|^4 - k^2 = 0 $

It follows that:

$$ \left(\bar \alpha |z|^2 - k \alpha\right) z = |z|^4 - k^2 \;\;\implies\;\; \left(\alpha |z|^2 - k \bar\alpha\right)\left(\bar \alpha |z|^2 - k \alpha\right) |z|^2 = \left(|z|^4 - k^2\right)^2 $$

The latter is a quartic in $\,|z|^2\,$, and the two real positive roots are the solutions for $\,|z|^2\,$.