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- Question 4 Exercise 8.2 @math-11-nbf:sol:unit08
- theta<\dfrac{\pi}{2}$, i.e. $\theta$ lies in QI. We have $$\sin\theta = \pm \sqrt{1-\cos^2}.$$ Since ... {7}}. \end{align*} (d) $\sin \dfrac{\theta}{2}$ We have $$\sin\left(\frac{\theta}{2} \right) = \pm \... {5}}} \end{align*} (e) $\cos \dfrac{\theta}{2}$ We have $$\cos\left(\frac{\theta}{2} \right) = \pm \... \frac{3\pi}{2}\), i.e., \(\theta\) lies in QIII. We have: \begin{align*} \sec \theta &= \pm \sqrt{1+
- Question 4, Exercise 1.3 @math-11-nbf:sol:unit01
- {7}{53}i\end{align} Put value of $\omega$ in (1), we have \begin{align} &(1-i) z+(1+i)\left(\dfrac{2}{... frac{155}{106}+\dfrac{145}{106}i\end{align} Thus, we have $$z=\dfrac{155}{106}+\dfrac{145}{106}i, \ome... ts(2) \end{align} Multiplying $(1)$ by $(1-2i)$, we get: \begin{align} &(1-2i)(2i z) + (1-2i)(3-2i) ... 3) \end{align} Multiplying equation (2) by $2i$, we get: \begin{align} &2i(1-2i) z + 2i(3+2i) \omega
- Question 9, Exercise 1.2 @math-11-nbf:sol:unit01
- **Solution.** Suppose $z=3 - \sqrt{-4}=3-2i$. We will use the following formulas: \[\text{Re}(z^{... \dfrac{7+2 i}{3-i}\right)^{-1}$. **Solution.** We use the following formulas: \[Re\left(\left(\fr... 2}.\] For \(z_1 = 7 + 2i\) and \(z_2 = 3 - i\), we have: \[x_1 = 7, \quad y_1 = 2, \quad x_2 = 3, ... frac{4+2 i}{2+5 i}\right)^{-2}$. **Solution.** We will use the following formulas: \begin{align}
- Question 5 Exercise 8.2 @math-11-nbf:sol:unit08
- $\sin 2\theta=\dfrac{24}{25}$, $2\theta$ in QII. We have $$\cos 2\theta = \pm \sqrt{1-\sin^2 2\theta}... frac{49}{625}} = -\frac{7}{25} \end{align*} Also we have $$\sin\theta = \pm \sqrt{\frac{1-\cos 2\thet... -\dfrac{7}{25}\) and \(2\theta\) lies in QIII. We have: \[\sin 2\theta = \pm \sqrt{1 - \cos^2 2\theta}\] Since \(2\theta\) lies in QIII, we know that \(\sin 2\theta < 0\). Therefore: \begin
- Question 10, Exercise 1.2 @math-11-nbf:sol:unit01
- \, -- (4) \end{align} From (1), (2), (3) and (4), we have: $$\left| z_1 \right| = \left| -z_1 \right| ... 10} + \frac{7}{10}i. \,\, -- (i) \end{align} Now, we have \begin{align} \overline{z_1} = -3 - 2i, \qua... + \frac{7}{10}i.\,\, -- (ii)$$ From (i) and (ii), we have \[ \overline{\left( \frac{z_1}{z_2} \right)}... n \[ z_1 = -3 + 2i, \quad z_2 = 1 - 3i. \] First we calculate \begin{align} z_1 z_2 &= (-3 + 2i)(1
- Question 11 and 12, Exercise 4.8 @math-11-nbf:sol:unit04
- \end{align*} Multiplying both sides by $k(k+2)$, we get \begin{align*} 1 = A(k+2) + Bk \ldots (2) \end{align*} Put $k=0$ in (2), we have \begin{align*} &1=2A + 0 \\ \implies & A = \... }. \end{align*} Put $k+2=0 \implies k=-2$ in (2), we have \begin{align*} &1=0-2B\\ \implies &B = -\fra... Using the values of $A$ and $B$ in equation (1), we get \begin{align*} \frac{1}{k(k+2)} &= \frac{1}{2
- Question 8, Exercise 1.2 @math-11-nbf:sol:unit01
- Solution.** Given: $$|2z-i|=4.$$ Put $z=x+i y$, we have \begin{align} & |2(x+iy)-i|=4 \\ \implies & ... n.** Given: $$|z-1|=|\bar{z}+i|.$$ Put $z=x+iy$, we have \begin{align} & |(x+iy)-1| = |(x-iy)+i| \\ \... qrt{x^2+(y-1)^2} \end{align} Squaring both sides, we get \begin{align} & (x-1)^2 + y^2 = x^2 + (y-1)^2... iven: $$|z-4i| + |z+4i| = 10.$$ Put $z = x + iy$, we have \begin{align} & |(x + iy) - 4i| + |(x + iy)
- Question 3, Exercise 1.4 @math-11-nbf:sol:unit01
- =\tan^{-1}\left(\dfrac{b}{a}\right). \end{align*} We can write these complex numbers in polar form as:... quad z=|z|e^{i\theta} \,\,-- (1) \end{align*} Now we have given \begin{align*} & \left(x_{1}+i y_{1}\r... z_3 \cdots z_n = z. \end{align*} By using $(1)$, we have \begin{align*} &|z_1| e^{i\theta_1}\cdot |z_... , -- (3) $$ Taking square on both sides of $(2)$, we get $$|z_1|^2\cdot|z_2|^2\cdot|z_3|^2\cdots |z_n|
- Question 2, Exercise 2.6 @math-11-nbf:sol:unit02
- _{2}+4 x_{3}=0\cdots(vi)\\ \end{align*} (iv)-(v), we have\\ \begin{align*} &\begin{array}{cccc} 2x_1&+... 3\\ \end{align*} Put the value of $x_2$ in (vi), we have \begin{align*} &3 x_{1}-2(\frac{11}{13})x_{3... lign*} If $\lambda=2$, put the value of $\lambda$ we have \begin{align*} x_{1} - 4x_{2} + 3x_{3} &= 0 ... = 0 \quad \text{(iii)} \end{align*} By using (i), we have \begin{align*} x_{1} &= 4x_{2} - 3x_{3} \q
- Question 20 and 21, Exercise 4.4 @math-11-nbf:sol:unit04
- \_\_\_ , \_\_\_ , \_\_\_ , 48$$ ** Solution. ** We have given $a_1=3$ and $a_5=48$. Assume $r$ be c... difference, then by general formula for nth term, we have $$ a_n=ar^{n-1}. $$ This gives \begin{align*... con="true"> **The good solution is as follows:** We have given $a_1=3$ and $a_5=48$. Assume $r$ be c... difference, then by general formula for nth term, we have $$ a_n=ar^{n-1}. $$ This gives \begin{align*
- Question 6 Exercise 8.2 @math-11-nbf:sol:unit08
- sin 15^{\circ} \cos 15^{\circ}$ ** Solution. ** We have double-angle identity: $$\sin 2 \theta = 2\s... 5^{\circ}-\sin ^{2} 15^{\circ}$ ** Solution. ** We have double-angle identity: $$\cos^2\theta -\sin^... ^{2}\left(\frac{\pi}{8}\right)$ ** Solution. ** We have a double-angle identity: $$\cos 2\alpha = 1-... pha=\cos 2\alpha.$$ Put $\alpha= \dfrac{\pi}{8}$, we have \begin{align*} 1-2\sin^2 \left(\frac{\pi}{8}
- Question 2, Exercise 1.2 @math-11-nbf:sol:unit01
- ltiplicative assocative law} \end{align} That is, we have proved $$(z_1 z_2)(z_3 z_4)=(z_1 z_3) (z_2 z... tiplicative associative law} \end{align} That is, we have proved $$(z_1 z_3) (z_2 z_4)=z_3 (z_1 z_2) z_4 ... (ii)$$ From (i) and (ii), we have the required result. **Remark:** For any three complex numbers $z_1$, $z_2$ and $z_3$, we have $$z_1 (z_2 z_3) = (z_1 z_2)z_3 = z_1 z_2 z_3
- Question 2, Exercise 2.3 @math-11-nbf:sol:unit02
- 1} = 3\), \(a_{12} = 2\), and \(a_{13} = 3\). Now we find their corresponding cofactors. \begin{align*... 5 \cdot 2) = (1) (4 - 10) = -6 \end{align*} Now, we use these cofactors to find the determinant: \beg... } = 2\), \(a_{12} = 3\), and \(a_{13} = -1\). Now we find their corresponding cofactors. \begin{align*... 1 - 0 \cdot 3) = (1) (-1) = -1 \end{align*} Now, we use these cofactors to find the determinant: \beg
- Question 20, 21 and 22, Exercise 4.3 @math-11-nbf:sol:unit04
- Given $a_{1}=7$, $a_{n}=139$, $S_{n}=876$. First we find $n$ and $d$.\\ As \begin{align} &S_n=\frac{n... implies & n=\frac{1752}{146}=12. \end{align} Also we have \begin{align} &a_n=a_1+(n-1)d\\ \implies & 1... ** Given $n=14$, $a_{n}=53$, $S_{n}=378$. First we find $a_1$ and $d$.\\ As \begin{align} &S_n=\frac... 7a_1=378-371 \\ \implies a_1=1. \end{align} Also we have \begin{align} &a_n=a_1+(n-1)d\\ \implies & 5
- Question 22 and 23, Exercise 4.4 @math-11-nbf:sol:unit04
- \_\_\_, \_\_\_, \dfrac{1}{4}$$ ** Solution. ** We have $a_1=8$ and $a_6=\frac{1}{4}$. Assume $r$ t... Then, by the general formula for the $n$th term, we have \\ $a_n = a_1 r^{n-1}.$\\ This gives\\ \begi... \\ \implies r &= \frac{1}{2}. \end{align*} Thus, we can find the missing terms: \begin{align*} a_2 &=... ric means. $$3 , \_\_\_ , 75$$ ** Solution. ** We have $a_1=3$ and $a_3=75$. Assume $r$ to be the