non zero complex numbers are orthogonal iff dot product is zero 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 
 A: 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 $z_{1} \cdot z_{2} = 0 \iff z_{1} \perp z_{2}$.
A: In the first half of your proof, you either write $icz_2$ when you should have
written $ic\bar z_2$, or you forgot to put a dot between the two $z_2$s.
In the second half, you forgot to conjugate $z_1$.
You therefore end up with a couple of sign errors, but only one of them matters.
Remember that only the real part (the part with the terms $x_1x_2$ and $y_1y_2$)
needs to be zero; the part with a factor of $i$ can be anything.
Then remember that you are trying to prove that the vectors corresponding to
these complex numbers are orthogonal. I assume you already know how to show that two
vectors are orthogonal.
You might want to do your "converse" part first and see if you find an equation in it
that can be made to do both the "if" and "only if" of the proof.
