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- Question 4, Exercise 1.3 @math-11-nbf:sol:unit01
- 1-i \end{align} \begin{align} \implies \omega & =\dfrac{1-i}{9+5i}\\ &=\dfrac{1-i}{9+5i}\times\dfrac{9-5i}{9-5i}\\ &=\dfrac{9-5-5i-9i}{81+25}\\ &=\dfrac{4-14i}{106}\\ &=\dfrac{2}{53}-\dfrac{7}{53}i\end{ali
- Question 2, Exercise 1.4 @math-11-nbf:sol:unit01
- n 2(i)===== Write the complex number $\left(\cos \dfrac{\pi}{6}+i \sin \dfrac{\pi}{6}\right)\left(\cos \dfrac{\pi}{3}+i \sin \dfrac{\pi}{3}\right)$ in rectangular form. ** Solution. ** Let $z_1=\cos \dfrac{\pi}{6
- Question 1, Exercise 9.1 @math-11-nbf:sol:unit09
- minimum values of the trigonometric function: $y=\dfrac{2}{3}-\dfrac{1}{2} \operatorname{Sin} \theta$ ** Solution. ** We know \begin{align*} -1 \leq \operator... n} \theta \leq 1 \end{align*} Multiplying with $-\dfrac{1}{2}$ \begin{align*} & \dfrac{1}{2} \geq -\dfrac{1}{2} \operatorname{Sin} \theta \geq -\dfrac{1}{2} \en
- Question 1, Exercise 8.1 @math-11-nbf:sol:unit08
- ign*} \begin{align*} \tan (\alpha + \beta) & = \dfrac{\tan\alpha + \tan\beta}{1-\tan\alpha \tan\beta} \\ \implies \tan (180 + 60) & = \dfrac{\tan 180 + \tan 60}{1-\tan 180 \tan 60} \\ \implies \tan (180 + 60) & = \dfrac{0 + \sqrt{3}}{1-(0)(\sqrt{3})} \\ & = \dfrac{\sqrt{3}}{1-0} = \sqrt{3} \end{align*} \begin{align*} \t
- Question 3, Exercise 1.1 @math-11-nbf:sol:unit01
- . ====Question 3(i)==== Simplify the following $\dfrac{(2+i)(3-2i)}{1+i}$ **Solution.** \begin{align}&\dfrac{(2+i)(3-2i)}{1+i}\\ =&\dfrac{6-2i^2+3i-4i}{1+i}\\ =&\dfrac{8-i}{1+i}\\ =&\dfrac{8-i}{1+i}\times \dfrac{1-i}{1-i}\\ =&\dfrac{8+i^2-8i-
- Question 7, Exercise 1.4 @math-11-nbf:sol:unit01
- llowing equation in Cartesian form: $\arg (z-1)=-\dfrac{\pi}{4}$ ** Solution. ** Suppose $z=x+iy$, as \begin{align*} &\arg (z-1)=-\dfrac{\pi}{4} \\ \implies & \arg(x+iy-1) = -\dfrac{\pi}{4} \\ \implies & \arg(x-1+iy) = -\dfrac{\pi}{4} \\ \implies & \tan^{-1}\left(\dfrac{y}{x-1}\right)
- Question 2, Exercise 8.1 @math-11-nbf:sol:unit08
- ht)\\ &= \cos 45 \cos 30 + \sin 45 \sin 30 \\ &= \dfrac{1}{\sqrt{2}}\cdot \dfrac{\sqrt{3}}{2} + \dfrac{1}{\sqrt{2}}\cdot \dfrac{1}{2} \\ & = \dfrac{\sqrt{3}}{2\sqrt{2}}+\dfrac{1}{2\sqrt{2}} \\ & = \dfrac{\sq
- Question 8, Exercise 1.4 @math-11-nbf:sol:unit01
- amplitude is $0.004 \mathrm{~mm}$ and angle is: $\dfrac{\pi}{4}$ ** Solution. ** Here we have $$x_{\max}=0.004, \quad \theta=\dfrac{\pi}{4}.$$ By using the formula \begin{align} x&=x_{\max} e^{i\theta} \\ &=0.004 e^{i\dfrac{\pi}{4}} \\ &=\frac{4}{1000} \left(\cos\left(\dfrac{\pi}{4}\right) +i \sin\left(\dfrac{\pi}{4}\right)\ri
- Question 7 and 8, Exercise 2.6 @math-11-nbf:sol:unit02
- & -11 \end{bmatrix}\\ A^{-1} &= \begin{bmatrix} \dfrac{-3}{62} & \dfrac{9}{62} & \dfrac{5}{62} \\ \dfrac{26}{62} & \dfrac{-16}{62} & \dfrac{-2}{62} \\ \dfrac{19}{62} & \dfrac{-5}{62} & \dfrac{
- Question 9, Exercise 8.1 @math-11-nbf:sol:unit08
- and $\beta$ are obtuse angles with $\sin \alpha=\dfrac{1}{\sqrt{2}}$ and $\cos \beta=-\dfrac{3}{5}$ find: $\sin (\alpha \pm \beta)$ ** Solution. ** Given: $\sin \alpha=\dfrac{1}{\sqrt{2}}$, $\alpha$ is obtuse angle, i.e. it is in QII.\\ $\cos \beta=-\dfrac{3}{5}$, $\beta$ is obtuse angle, i.e. it is in QI
- Question 3, Exercise 1.3 @math-11-nbf:sol:unit01
- Question 3(i)==== Solve the quadratic equation: $\dfrac{1}{3} z^{2}+2 z-16=0$. **Solution.** Given \begin{align}&\dfrac{1}{3}z^{2}+2 z-16=0\\ \implies &z^{2} + 6z - 48 =... \end{align} Apply the quadratic formula: $$ z = \dfrac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a},$$ where $$a = 1... t{and}\quad c = -48.$$ Then \begin{align} z& = \dfrac{{-6 \pm \sqrt{36-4(1)(-48)}}}{2 \cdot 1} \\ & = \
- Question 4 Exercise 8.2 @math-11-nbf:sol:unit08
- ac{\pi}{2}$ ** Solution. ** Given: $\cos\theta=\dfrac{3}{5}$ where $0<\theta<\dfrac{\pi}{2}$, i.e. $\theta$ lies in QI. We have $$\sin\theta = \pm \sqrt{1-\c... theta = -\frac{24}{7}}. \end{align*} (d) $\sin \dfrac{\theta}{2}$ We have $$\sin\left(\frac{\theta}{2}... pm \sqrt{\frac{1-\cos\theta}{2}}$$ As $0<\theta<\dfrac{\pi}{2}$ implies $0<\dfrac{\theta}{2}<\dfrac{\pi}
- Question 1, Exercise 2.2 @math-11-nbf:sol:unit02
- ht]$ of order $2 \times 2$ for which is $a_{i j}=\dfrac{i+3 j}{2}$ ** Solution. ** Given \( a_{ij} = \dfrac{i + 3j}{2} \). For \( i = 1, j = 1 \): \[ a_{11} = \dfrac{1 + 3 \cdot 1}{2} = \dfrac{1 + 3}{2} = \dfrac{4}{2} = 2 \] For \( i = 1, j = 2 \): \[ a_{12} = \dfrac{
- Question 2, Exercise 1.3 @math-11-nbf:sol:unit01
- )==== Solve the equation by completing square: $-\dfrac{1}{2} z^{2}-5 z+2=0$. **Solution.** \begin{align} -\dfrac{1}{2} z^{2} - 5z + 2& = 0 \end{align} Multiply th... n.** \begin{align} 4z^{2} + 5z &= 14\\ z^{2} + \dfrac{5}{4}z& = \dfrac{14}{4} \\ (z + \dfrac{5}{8})^2 - (\dfrac{5}{8})^2 &=\dfrac{7}{2} \\ (z + \dfrac{5}{8})^
- Question 7, Exercise 8.1 @math-11-nbf:sol:unit08
- $ and $\beta$ are acute angles with $\sin \alpha=\dfrac{12}{13}$ and $\tan \beta=\dfrac{4}{3}$ find \\ (i) $\sin(\alpha+\beta)$ (ii) $\cos(\alpha+\beta)$ (iii) ... a+\beta)$. ** Solution. ** Given: $\sin \alpha=\dfrac{12}{13}$, where $\alpha$ is acute angle, i.e. is in QI.\\ $\tan \beta=\dfrac{4}{3}$, where $\beta$ is acute angle, i.e. is in