If z = sqrt{2} - isqrt{2} is rotated through | vedclass

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(A) The given complex number is $z = \sqrt{2} - i\sqrt{2}$.Rotating a complex number $z$ by an angle $\alpha$ in the anti-clockwise direction about th

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$(2, 0)$

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(A) The given complex number is $z = \sqrt{2} - i\sqrt{2}$.Rotating a complex number $z$ by an angle $\alpha$ in the anti-clockwise direction about the origin is equivalent to multiplying $z$ by $e^{i\alpha}$.Here,$\alpha = 45^{\circ} = \frac{\pi}{4}$.Let the new position be $z_1 = z \cdot e^{i\pi/4}$.$z_1 = (\sqrt{2} - i\sqrt{2}) \cdot (\cos 45^{\circ} + i \sin 45^{\circ})$.$z_1 = (\sqrt{2} - i\sqrt{2}) \cdot (\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}})$.$z_1 = \sqrt{2}(\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}) - i\sqrt{2}(\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}})$.$z_1 = (1 + i) - (i - 1) = 1 + i - i + 1 = 2$.Thus,$z_1 = 2 + 0i$,which corresponds to the coordinates $(2, 0)$.

If $z = \sqrt{2} - i\sqrt{2}$ is rotated through an angle $45^{\circ}$ in the anti-clockwise direction about the origin,then the coordinates of its new position are

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