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Exercise 2.8 · Q9

Q.If z=2−2iz=2-2i, find the rotation of zz by θ\theta radians in the counter clockwise direction about the origin when

(i) θ=π3\theta=\dfrac\pi3
(ii) θ=2π3\theta=\dfrac{2\pi}3
(iii) θ=3π2\theta=\dfrac{3\pi}2.
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The rotation of zz by θ\theta radians counter-clockwise about the origin is z⋅cis⁡θz\cdot\operatorname{cis}\theta (§2.8.2, Note (ii)); we first put z=2−2iz=2-2i in polar form, then add each given θ\theta to its argument and convert back to rectangular form.

Step 0. A note on the printed question. In the source text, parts (ii) and (iii) show a corrupted symbol in place of θ\theta (a stray tab character rendered as "\theta"); read in context (rotating the same z=2−2iz=2-2i through three progressively larger angles), the intended values are θ=2π3\theta=\dfrac{2\pi}3 for (ii) and θ=3π2\theta=\dfrac{3\pi}2 for (iii).

Step 1. Write z=2−2iz=2-2i in polar form. x=2,y=−2⇒r=4+4=22x=2,y=-2\Rightarrow r=\sqrt{4+4}=2\sqrt2; reference angle α=tan⁡−1∣−22∣=π4\alpha=\tan^{-1}\left|\frac{-2}2\right|=\frac\pi4; Quadrant IV ⇒\Rightarrow argument =−π4=-\dfrac\pi4. So

z=22cis⁡(−π4).z=2\sqrt2\operatorname{cis}\left(-\frac\pi4\right).

Step 2. Rotation formula. The rotation of zz by θ\theta is z⋅cis⁡θ=22cis⁡(θ−π4)z\cdot\operatorname{cis}\theta=2\sqrt2\operatorname{cis}\left(\theta-\frac\pi4\right).

Step 3. Part (i): θ=π3\theta=\dfrac\pi3. New angle =π3−π4=4π−3π12=π12=15∘=\dfrac\pi3-\dfrac\pi4=\dfrac{4\pi-3\pi}{12}=\dfrac\pi{12}=15^\circ. Using cos⁡15∘=6+24, sin⁡15∘=6−24\cos15^\circ=\dfrac{\sqrt6+\sqrt2}4,\ \sin15^\circ=\dfrac{\sqrt6-\sqrt2}4:

z′=22(6+24+i 6−24)=2(6+2)2+i 2(6−2)2=23+22+i 23−22z'=2\sqrt2\left(\frac{\sqrt6+\sqrt2}4+i\,\frac{\sqrt6-\sqrt2}4\right)=\frac{\sqrt2(\sqrt6+\sqrt2)}2+i\,\frac{\sqrt2(\sqrt6-\sqrt2)}2=\frac{2\sqrt3+2}2+i\,\frac{2\sqrt3-2}2

=(1+3)+(3−1)i.=(1+\sqrt3)+(\sqrt3-1)i. …

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