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- Question 9, Exercise 1.2
- and imaginary parts of $\left(\dfrac{7+2 i}{3-i}\right)^{-1}$. **Solution.** We use the following for... \[Re\left(\left(\frac{x_1 + i y_1}{x_2 + i y_2}\right)^{-1}\right) = \frac{x_1 x_2 + y_1 y_2}{x_1^2 + y_1^2},\] \[Im\left(\left(\frac{x_1 + i y_1}{x_2 + i y_2}\right)^{-1}\right) = \frac{x_1 y_2 - x_2 y_1}{x_1^2 + y
- Question 7, Exercise 1.4
- i}{4} \\ \implies & \tan^{-1}\left(\dfrac{y}{x-1}\right) = -\dfrac{\pi}{4} \\ \implies & \dfrac{y}{x-1} = \tan\left(-\dfrac{\pi}{4}\right) \\ \implies & \dfrac{y}{x-1} = -1 \\ \implies & ... in Cartesian form: $z \bar{z}=4\left|e^{i \theta}\right|$ ** Solution. ** Suppose $z=x+iy$, then $\bar{... As \begin{align*} &z \bar{z}=4\left|e^{i \theta}\right| \\ \implies & (x+iy)(x-iy)=4|\cos\theta+i\sin\th
- Question 10, Exercise 1.2
- (i)==== For $z_{1}=-3+2 i$, verify: $$\left|z_{1}\right|=\left|-z_{1}\right|=\left|\overline{z_{!}}\right|=\left|-\overline{z_{!}}\right|.$$ **Solution.** \begin{align} |z_1| &= \sqrt{(-3)^2 + (2)^2} \\ &= \
- Question 3, Exercise 1.4
- =====Question 3(i)===== If $\left(x_{1}+i y_{1}\right)\left(x_{2}+i y_{2}\right)\left(x_{3}+i y_{3}\right) \ldots\left(x_{n}+i y_{n}\right)=a+i b$, show that: (i) $\left(x_{1}^{2}+y_{1}^{2}\right)\left(x_{2}^{
- Question 2, Exercise 1.4
- $\left(\cos \dfrac{\pi}{6}+i \sin \dfrac{\pi}{6}\right)\left(\cos \dfrac{\pi}{3}+i \sin \dfrac{\pi}{3}\right)$ in rectangular form. ** Solution. ** Let $z_1=... {3}} \\ & = e^{i\left(\frac{\pi}{6}+\frac{\pi}{3}\right)} \\ & = e^{i\frac{\pi}{2}} \\ & = \cos \dfrac{\p... $ \left(\cos \dfrac{\pi}{6}+i \sin \dfrac{\pi}{6}\right)\left(\cos \dfrac{\pi}{3}+i \sin \dfrac{\pi}{3}\r
- Question 6(i-ix), Exercise 1.4
- \sqrt{2}\left(\cos 315^{\circ}+i \sin 315^{\circ}\right)$ ** Solution. ** \begin{align} &\sqrt{2}\left(\cos 315^{\circ}+i \sin 315^{\circ}\right) \\ =& \sqrt{2} \left(\dfrac{1}{\sqrt{2}}-\dfrac{i}{\sqrt{2}} \right) \\ =& 1-i. \end{align} =====Question 6(ii)===== ... form: $5\left(\cos 210^{\circ}+i \sin 210^{\circ}\right)$ ** Solution. ** \begin{align*} &5\left(\cos 2
- Question 1, Exercise 1.4
- egin{align} \alpha & = \tan^{-1}\left|\frac{y}{x}\right| = \tan^{-1}\left|\frac{2\sqrt{3}}{2}\right|\\ & = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}. \end{align} Si... \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right). \] GOOD =====Question 1(ii)===== Write the fol... begin{align} \alpha &= \tan^{-1}\left|\frac{y}{x}\right| = \tan^{-1}\left|\frac{-\sqrt{3}}{3}\right| \\ &
- Question 4, Exercise 1.1
- 2+i)}=\dfrac{(1-5i)(2-i)+y(3-2i)}{(3-2i)(2-i)}\\ \Rightarrow \quad & \dfrac{x}{(2+i)}=\dfrac{(2+5i^2-10i-i)+(3y-2yi)}{(6+2i^2-4i-3i)}\\ \Rightarrow \quad & \dfrac{x}{(2+i)}=\dfrac{(2-5-11i)+(3y-2yi)}{(6+2i^2-4i-3i)}\\ \Rightarrow \quad & \dfrac{x}{(2+i)}=\dfrac{(3y-3)+(-11-2y)i}{(4-7i)}\\ \Rightarrow \quad & x(4-7i)=((3y-3)+(-11-2y)i)(2+i)\\ \R
- Question 1, Exercise 1.1
- }&=i\cdot{{i}^{30}}\\ &=i\cdot{{\left( {{i}^{2}} \right)}^{15}}\\ &=i\cdot{{\left( -1 \right)}^{15}} \quad \because i^2=-1\\ &=i\cdot(-1)\\ &=-i.\end{align} GO... ====Question 1(ii)==== Evaulate ${{\left( -i \right)}^{6}}$. **Solution.** \begin{align} {{\left( -i \right)}^{23}}&=(-1)^{23} i^{23} \\ &=-1 \cdot i^{22} \c
- Question 2, Exercise 1.2
- omplex numbers to prove that $$ \left(z_{1} z_{2}\right)\left(z_{3} z_{4}\right)=\left(z_{1} z_{3}\right)\left(z_{2} z_{4}\right)=z_{3}\left(z_{1} z_{2}\right) z_{4} $$ **Solution.** \begin{align} &(z_1 z_2
- Question 6(x-xvii), Exercise 1.4
- eft(\cos \dfrac{5 \pi}{4}+i \sin \dfrac{5 \pi}{4}\right)$ ** Solution. ** //Do yourself as previous par... eft(\cos \dfrac{7 \pi}{4}+i \sin \dfrac{7 \pi}{4}\right)$ ** Solution. ** //Do yourself as previous p... 2\left(\cos\dfrac{5\pi}{2}+i \sin \dfrac{5\pi}{2}\right)$ ** Solution. ** //Do yourself as previous part... }}\left(\cos \dfrac{\pi}{4}+i \sin \dfrac{\pi}{4}\right)$ ** Solution. ** //Do yourself as previous par
- Question 8, Exercise 1.2
- es: \begin{align} & \left(5\sqrt{x^2 + (y + 4)^2}\right)^2 = \left(25 + 4y\right)^2 \\ \implies & 25 \left(x^2 + (y + 4)^2\right) = 625 + 200y + 16y^2. \\ \implies & 25 \left(x^2 + y^2 + 8y + 16\right) = 625 + 200y + 16y^2 \\ \implies & 25x^2 + 25y^
- Question 4, Exercise 1.3
- &(1-i) z+(1+i)\left(\dfrac{2}{53}-\dfrac{7}{53}i \right)=3\\ \implies &(1-i) z+\dfrac{2}{53}+\dfrac{7}{53... (3-2i)\left( \dfrac{199}{205} + \dfrac{177}{205}i\right) = 1 + i \\ \implies & 2iz + \dfrac{597}{205}+\df... \dfrac{i}{2} z+\left(\dfrac{3}{4}-\dfrac{1}{2} i\right) \omega=\left(\dfrac{1}{2}+2 i\right)$. ** Solution. ** Given: \begin{align} &\dfrac{3}{i} z - (6 + 2i)
- Question 1, Exercise 1.3
- {3}{25} \\ = & z^2 - \left(\dfrac{\sqrt{3}}{5} i \right)^2 \\ = & \left(z + \dfrac{\sqrt{3}}{5}i\right)\left(z - \dfrac{\sqrt{3}}{5}i\right) \end{align} ====Question 1(v)==== Factorize the polynomial into li... z-15.$$ Since \begin{align}P\left(-\dfrac{3}{2} \right) &= 2\left(-\dfrac{3}{2}\right)^3 + 3\left(-\dfra
- Question 1, Review Exercise
- 1}$ is * (a) $\left(\frac{1}{3}, \frac{1}{4}\right)$ * (b) $\left(-\frac{1}{3},-\frac{1}{4}\right)$ * %%(c)%% $\left(\frac{3}{25}, \frac{-4}{25}\right)$ * (d) $\left(\frac{3}{25}, \frac{-4}{25}\right)$ \\ <btn type="link" collapse="a8">See Answer</bt