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in complex plane we have a line passing through $z_1$ and $z_2$. I want to find a line making an angle $\theta$ with this line and passing through $z_1$. How do I do this?

I know i can convert it to Cartesian by splitting it into real and imaginary parts and applying the rotation matrix. But I feel it might be easier in doing it directly in complex plane since rotation there is only a multiplication by $e^{i\theta}$ or $e^{-i\theta}$ depending on anticlockwise or clockwise direction.

Can someone please help me solve this completely and get the final expression? Regards

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One line is the set of points $\{z_1+t(z_2-z_1)e^{i\theta}: t \in \mathbb R\}$ and the other is $\{z_1+t(z_2-z_1)e^{-i\theta}:t \in \mathbb R\}$. [Since the original line is along $z_2-z_1$ you only have to rotate this vector].

You can also write the equation to the first line as $\frac {z-z_1} {|z-z_1|}=\frac {{(z_2-z_1)}e^{i\theta}} {|z_2-z_1|}$.

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  • $\begingroup$ thanks, is there also an expression which does not involve t and have z (a complex number) as a variable? Just like we have x, y as variables in coordinate system. $\endgroup$
    – maik
    Feb 28, 2019 at 6:20
  • $\begingroup$ i thought the expression of the line can also be written as $\frac{z-z_1}{|z-z_1|} = \frac{z_2-z_1}{|z_2-z_1|}$. And if that is so how do i apply rotation here? $\endgroup$
    – maik
    Feb 28, 2019 at 6:28
  • $\begingroup$ @maik I have now written the equation in the form you need. $\endgroup$ Feb 28, 2019 at 6:33
  • $\begingroup$ Thanks @kavi, a last request, does the parameter $t$ in your first expression has a physical significance? maybe it denotes the distance from $z_1$ or something on those lines $\endgroup$
    – maik
    Feb 28, 2019 at 6:35
  • $\begingroup$ It is the distance from a general point $z$ on the line to $z_1$ divided by the distance between $z_1$ and $z_2$ (with a sign depending on the side of $z_1$). $\endgroup$ Feb 28, 2019 at 6:39

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