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- Question 1, Exercise 2.6 @math-11-nbf:sol:unit02
- em of homogeneous linear equation for non-trivial solution if exists\\ $ 2 x_{1}-3 x_{2}+4 x_{3}=0$\\ $x_{1}... _{2}+3 x_{3}=0$\\ $4 x_{1}+x_{2}-6 x_{3}=0$\\ ** Solution. ** \begin{align*} &2 x_{1}-3 x_{2}+4 x_{3}=0\cdo... 4+36=0 \end{align*} So the system has non-trivial solution. \text{By}\quad(i)-2(ii), we have \begin{align*... es of $x_3$, there are infinite solutions. Hence solution is; \begin{align*} \left[ \begin{array}{c} x_3
- Question 1, Exercise 1.3 @math-11-nbf:sol:unit01
- polynomial into linear functions: $z^{2}+169$. **Solution.** \begin{align} & z^{2} + 169 \\ = & z^{2} - (... olynomial into linear functions: $2 z^{2}+18$. **Solution.** \begin{align} & 2z^2 + 18 \\ = &2(z^2 - (3i)^... nomial into linear functions: $3 z^{2}+363$. **Solution.** \begin{align} & 3z^2 + 363 \\ = & 3(z^2 - (1... into linear functions: $z^{2}+\dfrac{3}{25}$. **Solution.** \begin{align} & z^2 + \dfrac{3}{25} \\ = & z^
- Question 6(i-ix), Exercise 1.4 @math-11-nbf:sol:unit01
- os 315^{\circ}+i \sin 315^{\circ}\right)$ ** Solution. ** \begin{align} &\sqrt{2}\left(\cos 315^{\circ... t(\cos 210^{\circ}+i \sin 210^{\circ}\right)$ ** Solution. ** \begin{align*} &5\left(\cos 210^\circ + i \s... c{3 \pi}{2}+i \sin \dfrac{3 \pi}{2}\right)$ ** Solution. ** \begin{align*} &2\left(\cos \frac{3\pi}{2} +... {5 \pi}{6}+i \sin \dfrac{5 \pi}{6}\right)$ ** Solution. ** \begin{align*} &4\left(\cos \frac{5\pi}{6} +
- Question 2, Exercise 1.1 @math-11-nbf:sol:unit01
- ex number in the form $x+iy$: $(3+2i)+(2+4i)$ ** Solution. ** \begin{align}&(3+i2)+(2+i4)\\ =&(3+2)+(i2+i... lex number in the form $x+iy$: $(4+3i)-(2+5i)$ **Solution.** \begin{align}&(4+3i)-(2+5i)\\ =&(4-2)+(3i-5i... lex number in the form $x+iy$: $(4+7i)+(4-7i)$ **Solution.** \begin{align} &(4+7i)+(4-7i)\\ =&(4+4)+(7i-7... lex number in the form $x+iy$: $(2+5i)-(2-5i)$ **Solution.** \begin{align} &(2+5i)-(2-5i)\\ =&(2-2)+(5i+5i
- Question 3, Exercise 1.3 @math-11-nbf:sol:unit01
- ratic equation: $\dfrac{1}{3} z^{2}+2 z-16=0$. **Solution.** Given \begin{align}&\dfrac{1}{3}z^{2}+2 z-16=0... 57}}}{2} \\ &= -3 \pm \sqrt{57} \end{align} Hence Solution set $=\{ -3 \pm \sqrt{57} \}$. ====Question 3(i... uadratic equation: $z^{2}-\frac{1}{2} z+17=0$. **Solution.** Given $$ z^{2} - \frac{1}{2}z + 17 = 0 $$ Usi... 1 \pm \sqrt{271}i}{4} \end{align} Therefore, the solution set is: $\left\{\dfrac{1 \pm \sqrt{271}i}{4}\righ
- Question 6(x-xvii), Exercise 1.4 @math-11-nbf:sol:unit01
- rac{5 \pi}{4}+i \sin \dfrac{5 \pi}{4}\right)$ ** Solution. ** //Do yourself as previous parts.// =====Ques... rac{7 \pi}{4}+i \sin \dfrac{7 \pi}{4}\right)$ ** Solution. ** //Do yourself as previous parts.// =====Q... \dfrac{5\pi}{2}+i \sin \dfrac{5\pi}{2}\right)$ ** Solution. ** //Do yourself as previous parts.// =====Ques... \dfrac{\pi}{4}+i \sin \dfrac{\pi}{4}\right)$ ** Solution. ** //Do yourself as previous parts.// =====Que
- Question 2, Exercise 2.6 @math-11-nbf:sol:unit02
- homogeneous linear equation may have non-trivial solution. Also solve the system for value of $\lambda$.\\ ... _{2}-x_{3}=0$\\ $3 x_{1}-2 x_{2}+4 x_{3}=0$\\ ** Solution. ** \begin{align*} &2 x_{1}-\lambda x_{2}+x_{3}=0... \\ \end{align*} Homogenous system has non-trivial solution, if \begin{align*} &\left| \begin{array}{ccc} 2 &... homogeneous linear equation may have non-trivial solution. Also solve the system for value of $\lambda$.\\
- Question 10, Exercise 8.1 @math-11-nbf:sol:unit08
- ft(\dfrac{\pi}{2}-\alpha\right)=\cos \alpha$ ** Solution. ** \begin{align*} L.H.S & = \sin \left(\frac{\p... === Verify: $\cos (\pi-\alpha)=-\cos \alpha$ ** Solution. ** \begin{align*} L.H.S & = \cos(\pi - \alpha) ... frac{1}{\sqrt{2}}(\cos \alpha-\sin \alpha)$ ** Solution. ** \begin{align*} L.H.S &= \cos \left(\alpha+\f... =\frac{\sqrt{2}}{2}(\cos \beta+\sin \beta)$ ** Solution. ** \begin{align*} L.H.S & = \cos \left(\alpha +
- Question 11, Exercise 8.1 @math-11-nbf:sol:unit08
- ht) \cos \left(270^{\circ}-\lambda\right)}=1$ ** Solution. ** \begin{align*} L.H.S & = \dfrac{\sin \left(1... ght)+\cos \left(90^{\circ}+\alpha\right)}=-1$ ** Solution. ** \begin{align*} L.H.S & = \frac{\sin \left(90... \sin (\alpha+\beta)}{\cos \alpha \cos \beta}$ ** Solution. ** \begin{align*} L.H.S & = \tan \alpha+\tan \b... ^{2} \alpha=\sin ^{2} \alpha-\sin ^{2} \beta$ ** Solution. ** \begin{align*} L.H.S & = \sin (\alpha + \bet
- Question 2, Exercise 1.3 @math-11-nbf:sol:unit01
- quation by completing square: $z^{2}-6 z+2=0$. **Solution.** \begin{align} & z^2 - 6z + 2 = 0 \\ \implies ... pleting square: $-\dfrac{1}{2} z^{2}-5 z+2=0$. **Solution.** \begin{align} -\dfrac{1}{2} z^{2} - 5z + 2& =... \implies z &= -5 \pm \sqrt{29} \end{align} Hence solution set$=\{-5 \pm \sqrt{29}\}$ ====Question 2(iii)=... uation by completing square: $4 z^{2}+5 z=14$. **Solution.** \begin{align} 4z^{2} + 5z &= 14\\ z^{2} + \d
- Question 3, Exercise 2.6 @math-11-nbf:sol:unit02
- x+3 y+4 z=2$\\ $2 x+y+z=5$\\ $3 x-2 y+z=-3$\\ ** Solution. ** Given the system of equations: \begin{align*}... x &= \frac{46}{19} \end{align*} Therefore, the solution to the system is: $$x = \frac{46}{19}, \quad y ... y+z=2$\\ $2 x+2 y+6 z=1$\\ $3 x-4 y-5 z=3$\\ ** Solution. ** Given the system of equations: \begin{align*}... h is inconsistent, the system of equations has no solution. =====Question 3(iii)===== Solve the system o
- Question 5, Exercise 2.6 @math-11-nbf:sol:unit02
- }+3 x_{3}=1$\\ $3 x_{1}-7 x_{2}+4 x_{3}=10$\\ ** Solution. ** The above system may be written as $A X=B$; w... 2}\\ &= \frac{104}{52} = 2 \end{align*} Thus, the solution set is $(3, 1, 2)$. =====Question 5(ii)===== S... {2}+2 x_{3}=1$\\ $8 x_{1}+x_{2}+4 x_{3}=-1$\\ ** Solution. ** The above system maybe written as $AX = B $, ... + 12}{42}\\ &= \frac{0}{42} = 0 \end{align*} The solution set for the given system of equations using Crame
- Question 8, Exercise 1.2 @math-11-nbf:sol:unit01
- $ in terms of $x$ and $y$ by taking $z=x+i y$. **Solution.** Given: $$|2z-i|=4.$$ Put $z=x+i y$, we have ... $ in terms of $x$ and $y$ by taking $z=x+i y$. **Solution.** Given: $$|z-1|=|\bar{z}+i|.$$ Put $z=x+iy$, w... $ in terms of $x$ and $y$ by taking $z=x+i y$. **Solution.** Given: $$|z-4i| + |z+4i| = 10.$$ Put $z = x ... $ in terms of $x$ and $y$ by taking $z=x+i y$. **Solution.** Given: $$\dfrac{1}{2} Re(i \bar{z}) = 4.$$ P
- Question 9, Exercise 1.2 @math-11-nbf:sol:unit01
- nd real and imaginary parts of $(2+4 i)^{-1}$. **Solution.** Suppose $z=2+4i$. \begin{align} Re(2+4i)^{-... and imaginary parts of $(3-\sqrt{-4})^{-2}$. **Solution.** Suppose $z=3 - \sqrt{-4}=3-2i$. We will us... ts of $\left(\dfrac{7+2 i}{3-i}\right)^{-1}$. **Solution.** We use the following formulas: \[Re\left(\... of $\left(\dfrac{4+2 i}{2+5 i}\right)^{-2}$. **Solution.** We will use the following formulas: \begin
- Question 10, Exercise 1.2 @math-11-nbf:sol:unit01
- _{!}}\right|=\left|-\overline{z_{!}}\right|.$$ **Solution.** \begin{align} |z_1| &= \sqrt{(-3)^2 + (2)^2} ... )}=\frac{\overline{z_{1}}}{\overline{z_{2}}}$. **Solution.** Given \[z_1 = -3 + 2i, \quad z_2 = 1 - 3i\] ... z_{2}}=\overline{z_{1}}\,\, \overline{z_{2}}$. **Solution.** Given \[ z_1 = -3 + 2i, \quad z_2 = 1 - 3i.... {1}+z_{2}}=\overline{z_{1}}+\overline{z_{2}}$. **Solution.** Given \[ z_1 = -3 + 2i, \quad z_2 = 1 - 3i.