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Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.5

Solution

Given:

z1=7(cos⁡23π12+isin⁡23π12),z2=11(cos⁡11π12+isin⁡11π12)z_1=7\left(\cos\frac{23\pi}{12}+i\sin\frac{23\pi}{12}\right),\quad z_2=11\left(\cos\frac{11\pi}{12}+i\sin\frac{11\pi}{12}\right)

First convert to rectangular form.

Convert z1z_1 to x+iyx+iy

23π12=2π−π12\frac{23\pi}{12}=2\pi-\frac{\pi}{12} cos⁡23π12=cos⁡π12=6+24,sin⁡23π12=−sin⁡π12=−6−24\cos\frac{23\pi}{12}=\cos\frac{\pi}{12}=\frac{\sqrt{6}+\sqrt{2}}{4},\quad \sin\frac{23\pi}{12}=-\sin\frac{\pi}{12}=-\frac{\sqrt{6}-\sqrt{2}}{4} z1=7(6+2)4−7(6−2)4iz_1=\frac{7(\sqrt{6}+\sqrt{2})}{4}-\frac{7(\sqrt{6}-\sqrt{2})}{4}i

Convert z2z_2 to x+iyx+iy

11π12=π−π12\frac{11\pi}{12}=\pi-\frac{\pi}{12} cos⁡11π12=−cos⁡π12=−6+24,sin⁡11π12=sin⁡π12=6−24\cos\frac{11\pi}{12}=-\cos\frac{\pi}{12}=-\frac{\sqrt{6}+\sqrt{2}}{4},\quad \sin\frac{11\pi}{12}=\sin\frac{\pi}{12}=\frac{\sqrt{6}-\sqrt{2}}{4} z2=−11(6+2)4+11(6−2)4iz_2=-\frac{11(\sqrt{6}+\sqrt{2})}{4}+\frac{11(\sqrt{6}-\sqrt{2})}{4}i

(i) z1+z2z_1+z_2

z1+z2=−(6+2)+(6−2)i\boxed{z_1+z_2=-(\sqrt{6}+\sqrt{2})+(\sqrt{6}-\sqrt{2})i}

(ii) z1−z2z_1-z_2

z1−z2=9(6+2)2−9(6−2)2i\boxed{z_1-z_2=\frac{9(\sqrt{6}+\sqrt{2})}{2}-\frac{9(\sqrt{6}-\sqrt{2})}{2}i}

(iii) z1⋅z2z_1\cdot z_2

Use multiplication in polar form:

z1z2=(7⋅11)(cos⁡(23π12+11π12)+isin⁡(23π12+11π12))z_1z_2=(7\cdot 11)\left(\cos\left(\frac{23\pi}{12}+\frac{11\pi}{12}\right)+i\sin\left(\frac{23\pi}{12}+\frac{11\pi}{12}\right)\right) 23π12+11π12=34π12=17π6=2π+5π6\frac{23\pi}{12}+\frac{11\pi}{12}=\frac{34\pi}{12}=\frac{17\pi}{6}=2\pi+\frac{5\pi}{6}

So:

z1z2=77(cos⁡5π6+isin⁡5π6)z_1z_2=77\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right) cos⁡5π6=−32,sin⁡5π6=12\cos\frac{5\pi}{6}=-\frac{\sqrt{3}}{2},\quad \sin\frac{5\pi}{6}=\frac{1}{2} z1z2=−7732+772i\boxed{z_1z_2=-\frac{77\sqrt{3}}{2}+\frac{77}{2}i}

(iv) z1z2\dfrac{z_1}{z_2}

Use division in polar form:

z1z2=711(cos⁡(23π12−11π12)+isin⁡(23π12−11π12))\frac{z_1}{z_2}=\frac{7}{11}\left(\cos\left(\frac{23\pi}{12}-\frac{11\pi}{12}\right)+i\sin\left(\frac{23\pi}{12}-\frac{11\pi}{12}\right)\right) 23π12−11π12=π\frac{23\pi}{12}-\frac{11\pi}{12}=\pi cos⁡π=−1,sin⁡π=0\cos\pi=-1,\quad \sin\pi=0 z1z2=−711+0i\boxed{\frac{z_1}{z_2}=-\frac{7}{11}+0i}
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