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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

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  • $\begingroup$ Pick up a complex number and multiply it by that and see what happens ! $\endgroup$
    – alkabary
    May 22 '15 at 2:31
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    $\begingroup$ You can put that complex number in polar form. It may help with the geometric interpretation of the problem. $\endgroup$
    – MathNewbie
    May 22 '15 at 2:32
  • $\begingroup$ Or even just plot the number $z = \frac{1-i}{\sqrt2}$ on the complex plane, and remember that is the result of multiplying $1$ by $z$. More generally, see Geometric interpretation of the multiplication of complex numbers? $\endgroup$
    – David K
    May 22 '15 at 2:36
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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.

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$$\frac{(1-i)}{\sqrt{2}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i=\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}i=$$

$$\left|\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}i\right|e^{\arg\left(\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}i\right)i}=$$

$$\sqrt{\left(\frac{1}{2}\sqrt{2}\right)^2+\left(\frac{1}{2}\sqrt{2}\right)^2}e^{-\tan^{-1}\left(\frac{\frac{1}{2}\sqrt{2}}{\frac{1}{2}\sqrt{2}}\right)i}=$$

$$\sqrt{\frac{1}{2}+\frac{1}{2}}e^{-\tan^{-1}(1)i}=e^{-\frac{\pi}{4}i}$$

$----$

$$\frac{(1-i)}{\sqrt{2}}=\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}i=e^{-\frac{\pi}{4}i}=\cos\left(-\frac{\pi}{4}\right)+\sin\left(-\frac{\pi}{4}\right)i$$

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