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- Question 1,Review Exercise @math-11-nbf:sol:unit09
- qrt{3}}$ \\ <btn type="link" collapse="a1">See Answer</btn><collapse id="a1" collapsed="true">%%(b)%%:... defined \\ <btn type="link" collapse="a2">See Answer</btn><collapse id="a2" collapsed="true">(a): $0$... 2}}{3}$ \\ <btn type="link" collapse="a3">See Answer</btn><collapse id="a3" collapsed="true">(c): $\f... \theta$ \\ <btn type="link" collapse="a4">See Answer</btn><collapse id="a4" collapsed="true">(a):$\co
- Question 4 Exercise 8.2 @math-11-nbf:sol:unit08
- theta<\dfrac{\pi}{2}$, 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 1, Review Exercise @math-11-nbf:sol:unit08
- {2}}{2}$ \\ <btn type="link" collapse="a1">See Answer</btn><collapse id="a1" collapsed="true">%%(a)%%:... t{3}+1}$ \\ <btn type="link" collapse="a2">See Answer</btn><collapse id="a2" collapsed="true">(b): $\f... sqrt{2}}$\\ <btn type="link" collapse="a3">See Answer</btn><collapse id="a3" collapsed="true">(c): $\f... s \theta$\\ <btn type="link" collapse="a4">See Answer</btn><collapse id="a4" collapsed="true">(d):$-\c
- 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
- MCQs: Math 11 NBF
- rational \\ <btn type="link" collapse="a1">See Answer</btn><collapse id="a1" collapsed="true">%%(c)%%:... (d) no \\ <btn type="link" collapse="a2">See Answer</btn><collapse id="a2" collapsed="true">(b): two... $(1,1)$ \\ <btn type="link" collapse="a3">See Answer</btn><collapse id="a3" collapsed="true">(a): $(0... \bar{z}$ \\ <btn type="link" collapse="a4">See Answer</btn><collapse id="a4" collapsed="true">(d): $\b
- Question 1, Review Exercise @math-11-nbf:sol:unit01
- rational \\ <btn type="link" collapse="a1">See Answer</btn><collapse id="a1" collapsed="true">%%(c)%%:... (d) no \\ <btn type="link" collapse="a2">See Answer</btn><collapse id="a2" collapsed="true">(b): two... $(1,1)$ \\ <btn type="link" collapse="a3">See Answer</btn><collapse id="a3" collapsed="true">(a): $(0... \bar{z}$ \\ <btn type="link" collapse="a4">See Answer</btn><collapse id="a4" collapsed="true">(d): $\b
- Question 1, Review Exercise @math-11-nbf:sol:unit02
- times n$ \\ <btn type="link" collapse="a1">See Answer</btn><collapse id="a1" collapsed="true">%%(b)%%:... imes n$ \\ <btn type="link" collapse="a2">See Answer</btn><collapse id="a2" collapsed="true">(d): $n ... _{i j}$ \\ <btn type="link" collapse="a3">See Answer</btn><collapse id="a3" collapsed="true">(d): $a_... of these \\ <btn type="link" collapse="a4">See Answer</btn><collapse id="a4" collapsed="true">(b): mul
- Question 1, Review Exercise @math-11-nbf:sol:unit05
- 1)(x-2)$ \\ <btn type="link" collapse="a1">See Answer</btn><collapse id="a1" collapsed="true">%%(d)%%:... (d) $3$ \\ <btn type="link" collapse="a2">See Answer</btn><collapse id="a2" collapsed="true">(a): $ 0... d) $ 3$ \\ <btn type="link" collapse="a3">See Answer</btn><collapse id="a3" collapsed="true">(b): $-3... $, then $x+1$ will be its:\\ * (a) divisor as well as factor\\ * (b) dividend\\ * %%(c)%%
- 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 3, Exercise 2.5 @math-11-nbf:sol:unit02
- {array}\right]$ if it exists. Also verify your answer by showing that $A A^{-1}=A^{-1} A=I$.\\ ** Sol... 2} \end{array} \right]\\ \end{align*} To verify, we need to show that \( A A^{-1} = A^{-1} A = I \): ... {array}\right]$ if it exists. Also verify your answer by showing that $A A^{-1}=A^{-1} A=I$.\\ ** Sol... 15-4)+5(-6)\\ &=-2+19-3-19\neq 0 \end{align*} Now we will find $A^{-1}$ \begin{align*} &\quad\left[ \b