1
$\begingroup$

I have two complex numbers, $z_1$ and $z_2$, that both have the modules equal to 1 and their arguments are $\theta_1$ and $\theta_2$, respectively.

I'd like to verify that $$\frac{(z_1+z_2)^2}{z_1\cdot z_2}$$ is a real number or equal to zero.

So we have $$z_1=\cos{\theta_1}+i\sin{\theta_1}\quad\text{and}\quad z_2=\cos{\theta_2}+i\sin{\theta_2}.$$ I figured out I should to something like this: $$\frac{(z_1+z_2)^2}{z_1\cdot z_2} \geq 0 \rightarrow \frac{z_1^2+2z_1z_2+z{_2}^2}{z_1\cdot z_2} \geq 0$$

I'm stuck and I think plugging in the values above in the inequality below isn't ideal. Any hints on what's the best approach?

$\endgroup$
  • $\begingroup$ In the last line, use the fact that $|z| = 1$ implies that $1/z = |z|/z = \overline{z}$. $\endgroup$ – anomaly Jul 12 '16 at 16:25
1
$\begingroup$

We get:

  • $$z_1\space\wedge\space z_2\in\mathbb{C}$$
  • $$|z_1|=|z_2|=1$$
  • $$\arg(z_1)=\theta_1$$
  • $$\arg(z_2)=\theta_2$$

Now, see that:

  • $$\left(z_1+z_2\right)^2=\left(|z_1|e^{\arg(z_1)i}+|z_2|e^{\arg(z_2)i}\right)^2=\left(e^{\theta_1i}+e^{\theta_2i}\right)^2=2e^{(\theta_1+\theta_2)i}+e^{2\theta_1i}+e^{2\theta_2i}$$
  • $$z_1\cdot z_2=|z_1|e^{\arg(z_1)i}\cdot|z_2|e^{\arg(z_2)i}=e^{\left(\theta_1+\theta_2\right)i}$$

So:

$$\frac{\left(z_1+z_2\right)^2}{z_1\cdot z_2}=\frac{2e^{(\theta_1+\theta_2)i}+e^{2\theta_1i}+e^{2\theta_2i}}{e^{\left(\theta_1+\theta_2\right)i}}=$$ $$\frac{2e^{(\theta_1+\theta_2)i}}{e^{\left(\theta_1+\theta_2\right)i}}+\frac{e^{2\theta_1i}}{e^{\left(\theta_1+\theta_2\right)i}}+\frac{e^{2\theta_2i}}{e^{\left(\theta_1+\theta_2\right)i}}=$$ $$2+\frac{e^{2\theta_1i}}{e^{\theta_1i}\cdot e^{\theta_2i}}+\frac{e^{2\theta_2i}}{e^{\theta_1i}\cdot e^{\theta_2i}}=$$ $$2+\frac{e^{\theta_1i}}{e^{\theta_2i}}+\frac{e^{\theta_2i}}{e^{\theta_1i}}=$$ $$2+e^{(\theta_1-\theta_2)i}+e^{(\theta_2-\theta_1)i}=$$ $$2+\cos(\theta_1-\theta_2)+\sin(\theta_1-\theta_2)i+\cos(\theta_2-\theta_1)+\sin(\theta_2-\theta_1)i=$$ $$2+2\cos(\theta_1-\theta_2)+0i=$$ $$2\left(1+\cos(\theta_1-\theta_2)\right)$$

$\endgroup$
2
$\begingroup$

To convert a complex ratio into something nicer you multiply both numerator and denominator by the conjugate of the denominator. This is particularly nice as the modulus of the two complex number is 1 and the algebra works out nicely :

$$\frac{(z_{1}+z_{2})^{2}}{z_{1}z_{2}} = \frac{(z_{1}+z_{2})^{2}\bar{z}_{1}\bar{z}_{2}}{|z_{1}z_{2}|^{2}} = |z_{1}|^{2}z_{1}\bar{z}_{2} + 2|z_{1}z_{2}|^{2} + |z_{2}|^{2}z_{2}\bar{z}_{1} = z_{1}\bar{z}_{2} + 2 + z_{2}\bar{z}_{1}$$

2 is obviously Real and $z_{1}\bar{z}_{2} + z_{2}\bar{z}_{1}$ is Real as it is invariant under conjugation. Job done.

$\endgroup$
1
$\begingroup$

$ arg (z_1 + z_2) = \theta_1/2 + \theta_2/2\\ arg (z_1 + z_2)^2 = \theta_1 + \theta_2\\ arg (z_1z_1) = \theta_1 + \theta_2\\ arg(\frac {(z_1 + z_2)^2}{z_1z_1}) = 0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.