This is the question I am stuck with. I tried to solve it as in image below.
I am not getting the correct answer. What mistake am I doing? Please help. Thank you
Thanks Adren, Thanks Nilabro for your answers. I want to know, can we solve this question by the same approach as I did? If yes, please tell how.
this rotation is decribed by :
$$R:z\mapsto e^{i\pi/4}(z-(1+2i))+1+2i$$
Hence :
$$R(3+4i)=\frac{\sqrt2}{2}(1+i)(2+2i)+1+2i=\boxed{1+2(1+\sqrt2)i}$$
It's worth to know that, in general, the rotation with angle $\theta$ (radians) around the point whose affix is $z_0$ is described by :
$$z\mapsto e^{i\theta}(z-z_0)+z_0$$
To rotate $z$ about $z_1$, shift the origin to $z_1$, then apply a rotation by $π/4$ radians counterclockwise about this new origin, and then shift the origin back to its original position.
In action:
$$ z' = (z - z_1)e^{\frac{iπ}{4}} + z_1. $$
Hence,
$ z' = (2 + 2i)\left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i\right) + 1 + 2i \\ \implies z' = \left({2\over \sqrt{2}} - {2\over \sqrt{2}} + 1\right) + i\left({2\over \sqrt{2}} + {2\over \sqrt{2}} + 2 \right) \\ \implies z' = 1 + (2 + 2\sqrt{2})i. $