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- Question 4, Exercise 1.3
- z-(2+5 i) \omega=2+3 i$. ** Solution. ** \begin{align} &(1-i) z+(1+i) \omega=3 \quad \cdots(1)\\ &2 z-(2+5 i) \omega=2+3i \quad\cdots(2) \end{align} Multiplying Eq. (1) by $2$: \begin{align} &(2-2i)z+(2+2i) \omega=6 \quad \cdots (3) \end{align} Multiplying Eq. (2) by $(1-i)$: \begin{align} &2
- Question 10, Exercise 1.2
- \overline{z_{!}}\right|.$$ **Solution.** \begin{align} |z_1| &= \sqrt{(-3)^2 + (2)^2} \\ &= \sqrt{9 + 4} = \sqrt{13} \,\, -- (1) \end{align} Now \begin{align} -z_1 &= -(-3 + 2i) = 3 - 2i\\ \implies |-z_1| &= \sqrt{(3)^2 + (-2)^2} \\ &= \sqrt{9 + 4} = \sqrt{13} \,\, -- (2) \end{align} Also \begin{align} \overline{z_1} &= -3 - 2i
- Question 1, Exercise 1.3
- ar functions: $z^{2}+169$. **Solution.** \begin{align} & z^{2} + 169 \\ = & z^{2} - (13i)^2 \\ = &(z + 13i)(z - 13i). \end{align} ====Question 1(ii)==== Factorize the polynomial ... r functions: $2 z^{2}+18$. **Solution.** \begin{align} & 2z^2 + 18 \\ = &2(z^2 - (3i)^2)\\ = &2(z + 3i)(z - 3i) \end{align} ====Question 1(iii)==== Factorize the polynomia
- Question 9, Exercise 1.2
- {-1}$. **Solution.** Suppose $z=2+4i$. \begin{align} Re(2+4i)^{-1} & = Re(z^{-1}) = \dfrac{Re(z)}{|z|... ^2+4^2} = \dfrac{2}{20}\\ &= \dfrac{1}{10}. \end{align} \begin{align} Im(2+4i)^{-1} & = Im(z^{-1}) = -\dfrac{Im(z)}{|z|^2} \\ & =-\dfrac{4}{2^2+4^2} = -\dfrac{4}{20}\\ &= \dfrac{1}{5}. \end{align} GOOD ====Question 9(ii)==== Find real and imagin
- Question 1, Exercise 1.4
- ** Let $z=x+iy=2 + i 2 \sqrt{3}$. We have \begin{align} r & = \sqrt{x^2 + y^2} = \sqrt{2^2 + (2\sqrt{3})^2} \\ & = \sqrt{4 + 12} = \sqrt{16} = 4. \end{align} and \begin{align} \alpha & = \tan^{-1}\left|\frac{y}{x}\right| = \tan^{-1}\left|\frac{2\sqrt{3}}{2}\rig... \\ & = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}. \end{align} Since the complex number \( 2 + i 2 \sqrt{3} \)
- Question 7, Exercise 1.4
- 4}$ ** Solution. ** Suppose $z=x+iy$, as \begin{align*} &\arg (z-1)=-\dfrac{\pi}{4} \\ \implies & \arg(... \ \implies & y = -x+1 \\ \implies & x+y = 1. \end{align*} As required. =====Question 7(ii)===== Convert ... Suppose $z=x+iy$, then $\bar{z}=x-iy$. As \begin{align*} &z \bar{z}=4\left|e^{i \theta}\right| \\ \impli... +y^2 = 4\sqrt{1} \\ \implies & x^2+y^2 = 4. \end{align*} =====Question 7(iii)===== Convert the following
- Question 4, Exercise 1.1
- ng: $(2+3i)x+(1+3i)y+2=0$ **Solution.** \begin{align}&(2+3i)x+(1+3i)y+2=0\\ \implies &(2x+y+2)+(3x+3y)i=0.\end{align} Comparing real and imaginary parts \begin{align} 2x+y+2&=0 \quad \cdots(1)\\ 3x+3y&=0\quad \cdots (2) \end{align} From (2), \begin{align} &3x=-3y \\ x=-y \quad .
- Question 8, Exercise 1.2
- iven: $$|2z-i|=4.$$ Put $z=x+i y$, we have \begin{align} & |2(x+iy)-i|=4 \\ \implies & |2x+i(2y-1)|=4 \\ \implies & \sqrt{(2x)^2+(2y-1)^2}=4 \end{align} Squaring on both sides \begin{align} & (2x)^2+(2y-1)^2 = 16\\ \implies & 4x^2+4y^2-4y+1-16=0 \\ \implies & 4x^2+4y^2-4y-15=0, \end{align} as required. GOOD ====Question 8(ii)==== Write $
- Question 2, Exercise 1.3
- g square: $z^{2}-6 z+2=0$. **Solution.** \begin{align} & z^2 - 6z + 2 = 0 \\ \implies & z^2 - 2(3)(z)+9... (z - 3)^2+7= 0 \\ \implies & (z - 3)^2 = 7. \end{align} Take the square root of both sides: \begin{align} &z - 3 = \pm \sqrt{7} \\ \implies &z = 3 \pm \sqrt{7}\end{align} Hence Solutioin set=$\{3 \pm \sqrt{7}\}$. ====
- Question 6(i-ix), Exercise 1.4
- 315^{\circ}\right)$ ** Solution. ** \begin{align} &\sqrt{2}\left(\cos 315^{\circ}+i \sin 315^{\cir... t{2}}-\dfrac{i}{\sqrt{2}} \right) \\ =& 1-i. \end{align} =====Question 6(ii)===== Write a given complex n... \sin 210^{\circ}\right)$ ** Solution. ** \begin{align*} &5\left(\cos 210^\circ + i \sin 210^\circ\right... t) \\ =& -\frac{5\sqrt{3}}{2} - \frac{5}{2}i \end{align*} =====Question 6(iii)===== Write a given complex
- Question 2, Exercise 1.1
- $x+iy$: $(3+2i)+(2+4i)$ ** Solution. ** \begin{align}&(3+i2)+(2+i4)\\ =&(3+2)+(i2+i4)\\ =&5+i6\end{align} GOOD ====Question 2(ii)==== Write the following co... m $x+iy$: $(4+3i)-(2+5i)$ **Solution.** \begin{align}&(4+3i)-(2+5i)\\ =&(4-2)+(3i-5i)\\ =&2-2i\end{align} GOOD ====Question 2(iii)==== Write the following co
- Question 3, Exercise 1.4
- r+iy_r$, $r=1,2,...,n$ and $z=a+ib$. Then \begin{align*} &|z_r|=\sqrt{x_r^2+y_r^2} \quad \text{and}\quad... d \theta=\tan^{-1}\left(\dfrac{b}{a}\right). \end{align*} We can write these complex numbers in polar form as: \begin{align*} z_r=|z_r| e^{i\theta_k} \quad \text{and}\quad z=|z|e^{i\theta} \,\,-- (1) \end{align*} Now we have given \begin{align*} & \left(x_{1}+
- Question 7, Exercise 1.1
- **Solution.** Suppose $$z=11+12i$$ Then \begin{align}|z|&= \sqrt{(11)^2+(12)^2}\\ &=\sqrt{265}\end{align} Hence $|11+12 i|=\sqrt{265}$. GOOD ====Question ... olution.** Suppose $z=(2+3i)ā(2+6i)$, then \begin{align}z&=2+3iā2ā6i\\ &=-3i \end{align} Now \begin{align} |z| &= \sqrt{0^2+(-3)^2} \\ &= \sqrt{9} = 3. \end{al
- Question 3, Exercise 1.1
- $\dfrac{(2+i)(3-2i)}{1+i}$ **Solution.** \begin{align}&\dfrac{(2+i)(3-2i)}{1+i}\\ =&\dfrac{6-2i^2+3i-4i... dfrac{7-9i}{2}\\ =&\dfrac{7}{2}-\dfrac{9}{2}i\end{align} GOOD ====Question 3(ii)==== Simplify the following $\dfrac{1+i}{(2+i)^2}$ **Solution.** \begin{align}&\dfrac{1+i}{(2+i)^2}\\ =&\dfrac{1+i}{4+i^2+4i}\\... c{7-i}{25}\\ =&\dfrac{7}{25}-\dfrac{1}{25}i. \end{align} GOOD ====Question 3(iii)==== Simplify the follo
- Question 3, Exercise 1.3
- }{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 = 0 \end{align} Apply the quadratic formula: $$ z = \dfrac{{-b \... = 6,\quad \text{and}\quad c = -48.$$ Then \begin{align} z& = \dfrac{{-6 \pm \sqrt{36-4(1)(-48)}}}{2 \cd... 6 \pm 2\sqrt{57}}}{2} \\ &= -3 \pm \sqrt{57} \end{align} Hence Solution set $=\{ -3 \pm \sqrt{57} \}$.