Recall that the dot (scalar) product of two planar vectors $v_1 = (x_1, y_1)$ and $v_2 = (x_2 , y_2)$ is given by $v_1. v_2 = x_1x_2 + y_1y_2$.
Exercise: Show that the vectors represented by the (nonzero) complex numbers $z_1$ and $z_2$ are orthogonal if and only if $z_1. z_2 = 0$
Note/Recall: that the dot product of the vectors represented by the complex numbers $z_1$ and $z_2$ is given by $z_1. z_2$ = Re$(\bar {z_1} z_2)$.
And orthogonality holds precisely when $z_1 = icz_2$ for some real $c$.
Attempt: Suppose that the vectors represented by the (nonzero) complex numbers $z_1$ and $z_2$ are orthogonal. Then $z_1 = icz_2$ for some real $c$. Then $z_1. z_2 = icz_2(z_2) = ic|z_2|^2$
Converse: Suppose $z_1. z_2 = 0$, then $z_1 . z_2 = (x_1 + iy_1)(x_2 + iy_2) = x_1x_2 + ix_1y_2 + ix_2y_1 – y_1y_2 = x_1x_2 – y_1y_2 + i(x_1y_2 + x_2y_1)$ must be equal to zero?
I don't know how to continue. Please can anyone please help me? Anything help/suggestion can help.
Thank you
So
Best Answer
You've defined $z_{1} \cdot z_{2} = Re(\overline{z_{1}} z_{2})$, and you want to show that $z_{1} \cdot z_{2} = 0 \iff z_{1} \perp z_{2}$. Without loss of generality, assume $\arg(z_{1}) \geq \arg(z_{2})$, and note that $z_{1} \perp z_{2}$ if, and only if, $\arg(z_{1}) - \arg(z_{2}) = \frac{\pi}{2}$ or $\arg(z_{1}) - \arg(z_{2}) = \frac{3\pi}{2}$.
Next, write the two complex numbers in polar form: $z_{1} = |z_{1}| e^{i \arg(z_{1})}$ and $z_{2} = |z_{2}| e^{i \arg(z_{2})}$. Then we have $$z_{1} \cdot z_{2} = |z_{1}||z_{2}| \cos(-\arg(z_{1}))\cos(\arg(z_{2}) - |z_{1}| |z_{2}| \sin(-\arg(z_{1})) \sin(\arg(z_{2})).$$ Because cosine is even and sine is odd, we have $$z_{1} \cdot z_{2} = |z_{1}||z_{2}| \cos(\arg(z_{1}))\cos(\arg(z_{2}) + |z_{1}| |z_{2}| \sin(\arg(z_{1})) \sin(\arg(z_{2})).$$ By trig identities, this is equal to $$z_{1} \cdot z_{2} = |z_{1}| |z_{2}| \cos(\arg(z_{1}) - \arg(z_{2})).$$ Since $|z_{1}| \neq 0$ and $|z_{2}| \neq 0$ (for, otherwise, $z_{1}$ and $z_{2}$ are trivially perpendicular), we must have $\cos(\arg(z_{1}) - \arg(z_{2})) = 0$, which can only happen if $\arg(z_{1}) - \arg(z_{2}) \in \{\frac{\pi}{2}, \frac{3\pi}{2}\}$.
For the converse, just work it backwards. We have $\arg(z_{1}) - \arg(z_{2})) \in \{\frac{\pi}{2}, \frac{3\pi}{2}\}$, so $z_{1} \cdot z_{2} = |z_{1}| |z_{2}| \cos( \arg(z_{1}) - \arg(z_{2})) = 0$.
So $z_{1} \cdot z_{2} = 0 \iff z_{1} \perp z_{2}$.