Rotating a complex number
Matthew Barrera
I have the complex number $3 + i$, and I am asked to get the complex number resulted by rotating the first one by $\frac{\pi}{4}$.
I got the polar form of the first one to get its angle ($18.43°$) but when I add $\frac{\pi}{4}$ to it and try to find its cartesian form, the result is different.
Am I doing something wrong with the procedure? If so, what can I do?
$\endgroup$ 12 Answers
$\begingroup$Rotating by $\frac\pi4$ is the same thing as multiplying your number by $\cos\left(\frac\pi4\right)+\sin\left(\frac\pi4\right)i=\frac1{\sqrt2}+\frac i{\sqrt2}$. So, the answer is$$\left(\frac1{\sqrt2}+\frac i{\sqrt2}\right)(3+i)=\sqrt2+2\sqrt2i,$$
$\endgroup$ 4 $\begingroup$Write in polar form: $re^{i\theta}$;
$r=\sqrt{10},$ $\tan(\theta) = 1/3$.
Rotate by $π/4$ counter clockwise.
$\tan(\theta') = \tan(\theta +π/4)=$
$\dfrac{\tan(\theta) + 1}{1-\tan(\theta)}= \dfrac{4/3}{2/3}=2;$
Rotated point: $ z' =re^{i\theta'}, $
$r=\sqrt{10}$ , $\tan(\theta') =2$.
Reverting to $z'= a' +ib':$
$z' = √2(1+2i).$
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