Verify $\frac{(z_1+z_2)^2}{z_1\cdot z_2} \geq 0$

Question

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?

Answer 1

We get:

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

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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:

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

Answer 2

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.

Answer 3

$ 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$

Answer 4

You can use the fact that a complex number $z$ is real if and only if $\bar{z}=z$. Note that $$\overline{\left( \frac{(z_1+z_2)^2}{z_1z_2} \right)} = \frac{(\overline{z_1}+\overline{z_2})^2}{\overline{z_1}\overline{z_2}} = \frac{\overline{z_1}^2+2\overline{z_1}\overline{z_2}+\overline{z_2}^2}{\overline{z_1}\overline{z_2}}=\frac{\overline{z_1}^2+2\overline{z_1}\overline{z_2}+\overline{z_2}^2}{\overline{z_1}\overline{z_2}}\cdot\frac{(z_1z_2)^2}{(z_1z_2)^2}=\frac{|z_1|^4(z_2)^2+2z_1z_2|z_1|^2|z_2|^2+|z_2|^4(z_1)^2}{(z_1z_2)|z_1|^2|z_2|^2}.$$ Since $|z_1|=|z_2|=1$, we have $$\overline{\left( \frac{(z_1+z_2)^2}{z_1z_2} \right)}=\frac{(z_2)^2+2z_1z_2+(z_1)^2}{z_1z_2}=\frac{(z_1+z_2)^2}{z_1z_2}.$$ Therefore, $\dfrac{(z_1+z_2)^2}{z_1z_2}$ is a real number.