How to express $z +\frac{1}{z}$ in polar form

Question

If we write $z=re^{i\theta}$ then we can express $z + \frac{1}{z}= \frac{r^{2}e^{i2\theta} +1}{r}. e^{-i\theta}$ Can we simplify it again and express it as $Re^{i \phi}$ ?

Answer 1

$$z+\frac{1}{z}=\left(r+\frac{1}{r}\right)\cos\theta+i\left(r-\frac{1}{r}\right)\sin\theta=$$ $$=\sqrt{\left(r+\frac{1}{r}\right)^2\cos^2\theta+\left(r-\frac{1}{r}\right)^2\sin^2\theta}e^{\left(\arctan\left(\frac{r^2-1}{r^2+1}\tan\theta\right)\right)i}=$$ $$=\sqrt{r^2+\frac{1}{r^2}+2\cos2\theta}e^{\left(\arctan\left(\frac{r^2-1}{r^2+1}\tan\theta\right)\right)i}.$$

Answer 2

Hint:

$$\omega =z+\dfrac{1}{z}=\biggl(r+\dfrac{1}{r}\biggr)\cos\theta+i\biggl(r-\dfrac{1}{r}\biggr)\sin\theta=Re^{i\phi}$$ $$R=\sqrt{\text{Re}^2(\omega)+\text{Im}^2(\omega)} \ \ \text{and} \ \phi =\arctan\biggl(\dfrac{\text{Im}(\omega)}{\text{Re}(\omega)}\biggr)$$

Can you proceed?

Answer 3

The expression is

$$\frac{z^2+1}{z}.$$

Its modulus is

$$R=\frac{|z^2+1|}{|z|}=\frac{\sqrt{(r^2\cos2\theta+1)^2+r^4\sin^22\theta}}r=\frac{\sqrt{r^4+2r^2\cos2\theta+1}}r$$

and its argument

$$\phi=\arctan\frac{r^2\sin2\theta}{r^2\cos2\theta+1}-\theta.$$

There is no further simplification.