Complex Numbers (Geometric Representations) What is the geometrical interpretation of this operation:
Multiplication by $\frac{\left(1-i\right)}{\sqrt{2}}$
Attempt:
multiplication by −i = rotate by −π/2
 A: First find the magnitude $r = \left|\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i\right| = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1$.
So multiplication by $\frac{\left(1-i\right)}{\sqrt{2}}$ does not change the magnitude.
Next find the argument by $\tan\theta = \frac{-1/\sqrt{2}}{1/\sqrt{2}} = -1 \implies \theta = -\frac{\pi}{4}$.
Thus multiplication by $\frac{\left(1-i\right)}{\sqrt{2}}$ represents a rotation by $\frac{\pi}{4}$ clockwise.
A: $$\frac{(1-i)}{\sqrt{2}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i=\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}i=$$
$$\left|\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}i\right|e^{\arg\left(\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}i\right)i}=$$
$$\sqrt{\left(\frac{1}{2}\sqrt{2}\right)^2+\left(\frac{1}{2}\sqrt{2}\right)^2}e^{-\tan^{-1}\left(\frac{\frac{1}{2}\sqrt{2}}{\frac{1}{2}\sqrt{2}}\right)i}=$$
$$\sqrt{\frac{1}{2}+\frac{1}{2}}e^{-\tan^{-1}(1)i}=e^{-\frac{\pi}{4}i}$$
$----$
$$\frac{(1-i)}{\sqrt{2}}=\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}i=e^{-\frac{\pi}{4}i}=\cos\left(-\frac{\pi}{4}\right)+\sin\left(-\frac{\pi}{4}\right)i$$
