# rotation in complex plane.

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

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 $$\frac {z-z_1} {|z-z_1|}=\frac {{(z_2-z_1)}e^{i\theta}} {|z_2-z_1|}$$.

• 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.
– maik
Feb 28, 2019 at 6:20
• 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?
– maik
Feb 28, 2019 at 6:28
• @maik I have now written the equation in the form you need. Feb 28, 2019 at 6:33
• 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
– maik
Feb 28, 2019 at 6:35
• 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 $z_1$). Feb 28, 2019 at 6:39