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Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.5

Solution

For a point (r,θ)(r,\theta):

  • If r>0r>0, move rr units at angle θ\theta.
  • If r<0r<0, move ∣r∣|r| units at angle θ+π\theta+\pi.

(i) (2,75∘)(2,75^\circ)

Here r=2>0r=2>0, so the point lies at angle 75∘75^\circ and distance 22 from the origin.

x=rcos⁡θ=2cos⁡75∘y=rsin⁡θ=2sin⁡75∘\begin{aligned} x&=r\cos\theta=2\cos 75^\circ\\ y&=r\sin\theta=2\sin 75^\circ \end{aligned}

(ii) (−3,120∘)(-3,120^\circ)

Here r=−3<0r=-3<0, so use angle 120∘+180∘=300∘120^\circ+180^\circ=300^\circ with distance 33.

x=rcos⁡θ=−3cos⁡120∘=32y=rsin⁡θ=−3sin⁡120∘=−332\begin{aligned} x&=r\cos\theta=-3\cos 120^\circ=\frac{3}{2}\\ y&=r\sin\theta=-3\sin 120^\circ=-\frac{3\sqrt{3}}{2} \end{aligned}

(iii) (2,π6)(2,\dfrac{\pi}{6})

Here θ=π6\theta=\dfrac{\pi}{6} and r=2>0r=2>0.

x=2cos⁡π6=3y=2sin⁡π6=1\begin{aligned} x&=2\cos\frac{\pi}{6}=\sqrt{3}\\ y&=2\sin\frac{\pi}{6}=1 \end{aligned}

(iv) (5,5π6)(5,\dfrac{5\pi}{6})

Here θ=5π6\theta=\dfrac{5\pi}{6} and r=5>0r=5>0.

x=5cos⁡5π6=−532y=5sin⁡5π6=52\begin{aligned} x&=5\cos\frac{5\pi}{6}=-\frac{5\sqrt{3}}{2}\\ y&=5\sin\frac{5\pi}{6}=\frac{5}{2} \end{aligned}

(v) (−52,π3)(-\dfrac{5}{2},\dfrac{\pi}{3})

Here r=−52<0r=-\dfrac{5}{2}<0, so use angle:

π3+π=4π3\frac{\pi}{3}+\pi=\frac{4\pi}{3} x=−52cos⁡π3=−54y=−52sin⁡π3=−534\begin{aligned} x&=-\frac{5}{2}\cos\frac{\pi}{3}=-\frac{5}{4}\\ y&=-\frac{5}{2}\sin\frac{\pi}{3}=-\frac{5\sqrt{3}}{4} \end{aligned}

(vi) (−3,−2π3)(-3,-\dfrac{2\pi}{3})

Here r=−3<0r=-3<0, so use angle:

−2π3+π=π3-\frac{2\pi}{3}+\pi=\frac{\pi}{3}

Also cos⁡(−2π3)=cos⁡(2π3)=−12\cos\left(-\frac{2\pi}{3}\right)=\cos\left(\frac{2\pi}{3}\right)=-\frac{1}{2} and sin⁡(−2π3)=−32\sin\left(-\frac{2\pi}{3}\right)=-\frac{\sqrt{3}}{2}.

x=−3cos⁡(−2π3)=32y=−3sin⁡(−2π3)=332\begin{aligned} x&=-3\cos\left(-\frac{2\pi}{3}\right)=\frac{3}{2}\\ y&=-3\sin\left(-\frac{2\pi}{3}\right)=\frac{3\sqrt{3}}{2} \end{aligned}

(vii) (92,19π12)(\dfrac{9}{2},\dfrac{19\pi}{12})

Here r=92>0r=\dfrac{9}{2}>0.

19π12=2π−5π12⇒cos⁡19π12=cos⁡5π12, sin⁡19π12=−sin⁡5π12\frac{19\pi}{12}=2\pi-\frac{5\pi}{12}\Rightarrow \cos\frac{19\pi}{12}=\cos\frac{5\pi}{12},\ \sin\frac{19\pi}{12}=-\sin\frac{5\pi}{12}

Use:

cos⁡5π12=6−24,sin⁡5π12=6+24\cos\frac{5\pi}{12}=\frac{\sqrt{6}-\sqrt{2}}{4},\quad \sin\frac{5\pi}{12}=\frac{\sqrt{6}+\sqrt{2}}{4} x=92cos⁡19π12=92⋅6−24=9(6−2)8y=92sin⁡19π12=92⋅(−6+24)=−9(6+2)8\begin{aligned} x&=\frac{9}{2}\cos\frac{19\pi}{12}=\frac{9}{2}\cdot\frac{\sqrt{6}-\sqrt{2}}{4}=\frac{9(\sqrt{6}-\sqrt{2})}{8}\\ y&=\frac{9}{2}\sin\frac{19\pi}{12}=\frac{9}{2}\cdot\left(-\frac{\sqrt{6}+\sqrt{2}}{4}\right)=-\frac{9(\sqrt{6}+\sqrt{2})}{8} \end{aligned}

(viii) (−52,5π12)(-\dfrac{5}{2},\dfrac{5\pi}{12})

Here r=−52<0r=-\dfrac{5}{2}<0, so use angle:

5π12+π=17π12\frac{5\pi}{12}+\pi=\frac{17\pi}{12}

Also:

cos⁡5π12=6−24,sin⁡5π12=6+24\cos\frac{5\pi}{12}=\frac{\sqrt{6}-\sqrt{2}}{4},\quad \sin\frac{5\pi}{12}=\frac{\sqrt{6}+\sqrt{2}}{4} x=−52cos⁡5π12=−5(6−2)8y=−52sin⁡5π12=−5(6+2)8\begin{aligned} x&=-\frac{5}{2}\cos\frac{5\pi}{12}=-\frac{5(\sqrt{6}-\sqrt{2})}{8}\\ y&=-\frac{5}{2}\sin\frac{5\pi}{12}=-\frac{5(\sqrt{6}+\sqrt{2})}{8} \end{aligned}
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