Answers to Calculus Practical 3

Question 1 §

$$-i=e^{\frac{3\pi }{2}}$$

$$-\frac{1}{2}-\frac{i}{2}=\sqrt{(-\frac{1}{2}{)}^2+(-\frac{1}{2}{)}^2}\cdot e^{i{\tan }^{-1}\left(1\right)}=\frac{1}{\sqrt{2}}\cdot e^{i\frac{5\pi }{4}}$$

$$-\sqrt{3}+i=\sqrt{(-\sqrt{3}{)}^2+(1{)}^2}\cdot e^{i{\tan }^{-1}\left(\frac{1}{-\sqrt{3}}\right)}=2e^{i\frac{5\pi }{6}}$$

$$3-\sqrt{3}i=\sqrt{(3{)}^2+(-\sqrt{3}{)}^2}\cdot e^{i{\tan }^{-1}\left(\frac{-\sqrt{3}}{3}\right)}=2\sqrt{3}\cdot e^{i\frac{11\pi }{6}}$$

$$-\frac{1}{\sqrt{3}}-i=\sqrt{(-\frac{1}{\sqrt{3}}{)}^2+(-1{)}^2}\cdot e^{i{\tan }^{-1}\left(\frac{-1}{-\frac{1}{\sqrt{3}}}\right)}=\frac{4}{3}{e}^{i\frac{4\pi }{3}}$$

Question 2 §

Plot these on the Argand diagram at home, using my drawing tablet

$$z^3=-i⟹\text{arg}(z)=\frac{\text{arg}(-i)+2k\pi }{3}=\frac{\frac{3\pi }{2}+2k\pi }{3}=\frac{(3+4k)\pi }{6}$$

$$\therefore z=e^{i\frac{3\pi }{6}},e^{i\frac{7\pi }{6}},e^{i\frac{11\pi }{6}}$$

---

$$z^3=8⟹\text{arg}(z)=\frac{\text{arg}(8)+2k\pi }{3}=\frac{2k\pi }{3}$$

$$\therefore z=1,e^{i\frac{2\pi }{3}},e^{i\frac{4\pi }{3}}$$

---

$$z^2=i⟹\text{arg}(z)=\frac{\text{arg}(i)+2k\pi }{2}=\frac{\frac{\pi }{2}+2k\pi }{2}=\frac{(1+4k)\pi }{4}$$

$$\therefore z=e^{i\frac{\pi }{4}},z=e^{i\frac{5\pi }{4}}$$

---

$$z^4=8\sqrt{2}(1-i)⟹\text{arg}(z)=\frac{\text{arg}(8\sqrt{2}(1-i))+2k\pi }{4}=\frac{\frac{7\pi }{4}+2k\pi }{4}=\frac{(7+8k)\pi }{16}$$

$$\therefore z=e^{i\frac{7\pi }{16}},e^{i\frac{15\pi }{16}},e^{i\frac{23\pi }{16}},e^{i\frac{31\pi }{16}}$$

Question 3 §

$$(\cos \theta +i\sin \theta {)}^3=\cos 3\theta +i\sin 3\theta$$

$$⟹\text{LHS}={\cos }^3\theta +i{\cos }^2\theta \sin \theta -\cos \theta {\sin }^2\theta -i{\sin }^3\theta ,\text{RHS}=\cos 3\theta +i\sin 3\theta$$

$$⟹\Re :{\cos }^3\theta -\cos \theta {\sin }^2\theta =\cos 3\theta ,\Im :{\cos }^2\theta \sin \theta -{\sin }^3\theta =\sin 3\theta$$

$$\therefore \sin 3\theta =\sin \theta -2{\sin }^3\theta ,\text{ by using }{\sin }^2\theta +{\cos }^2\theta \equiv 1,$$

$$\cos 3\theta =2{\cos }^3\theta -\cos \theta ,\text{ by using }{\sin }^2\theta +{\cos }^2\theta \equiv 1.$$

Question 4 §

I’m assuming there’s an error in the question, where an $i$ is missed from the original given equation, given this:

$$(\sqrt{3}-i{)}^{10}=(2e^{i\frac{11\pi }{6}}{)}^{10}=2^{10}\cdot e^{i\frac{110\pi }{6}}=2^{10}\cdot e^{i\frac{\pi }{3}}$$

$$⟹\sqrt{a^2+b^2}=2^{10},\frac{a}{b}=\tan \frac{\pi }{3}=\sqrt{3}:z=a+bi$$

$$⟹2^{20}=(b\sqrt{3}{)}^2+b^2$$

$$⟹2^9=b,a=2^9\sqrt{3}$$

$$\therefore 2^9(\sqrt{3}+i)\text{ or }{2}^9\sqrt{3}+2^{9}i$$

This seems very incorrect - check later

Question 5 §

Given that ${\omega }^0=1,{\omega }^1=e^{i\frac{2\pi }{3}},{\omega }^2=e^{i\frac{4\pi }{3}}$, show that ${\omega }^3=1$ and that $1+\omega +{\omega }^2=0$.

$${\omega }^3={\omega }^2\cdot {\omega }^1=e^{i\frac{4\pi }{3}}\cdot e^{i\frac{2\pi }{3}}=e^{i2\pi }=e^{i0}=e^0=1■$$

$$1+\omega +{\omega }^2=e^{i0}+e^{i\frac{4\pi }{3}}+e^{i\frac{2\pi }{3}}$$

Converting each complex number into the form $a+bi$ gives the following:

$$e^{i0}:\tan 0=\frac{a_1}{b_1},1^2=a_1^2+b_1^2\therefore a_1=0,b_1=\pm 1$$

$$e^{i\frac{4\pi }{3}}:\tan \frac{4\pi }{3}=\frac{a_2}{b_2},1^2=a_2^2+b_2^2\therefore a_1=\pm \frac{\sqrt{3}}{2},b_2=\pm \frac{1}{2}$$

$$e^{i\frac{2\pi }{3}}:\tan \frac{2\pi }{3}=\frac{a_3}{b_3},1^2=a_3^2+b_3^2\therefore a_3=$$