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- Question 4, Exercise 1.3
- ots(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{ali... lies & (2-2i)z-(7+3i)\omega=5+i \quad \cdots (4) \end{align} $(3)-(4)$ implies \begin{align} (9+5i) \omega=1-i \end{align} \begin{align} \implies \omega & =\dfrac{1-
- Question 10, Exercise 1.2
- )^2} \\ &= \sqrt{9 + 4} = \sqrt{13} \,\, -- (1) \end{align} Now \begin{align} -z_1 &= -(-3 + 2i) =... 2)^2} \\ &= \sqrt{9 + 4} = \sqrt{13} \,\, -- (2) \end{align} Also \begin{align} \overline{z_1} &= -3... ^2} \\ & = \sqrt{9 + 4} = \sqrt{13} \,\, -- (3) \end{align} Now \begin{align} -\overline{z_1} &= -(... )^2} \\ & = \sqrt{9 + 4} = \sqrt{13} \,\, -- (4) \end{align} From (1), (2), (3) and (4), we have: $$\le
- Question 1, Exercise 1.3
- \\ = & z^{2} - (13i)^2 \\ = &(z + 13i)(z - 13i). \end{align} ====Question 1(ii)==== Factorize the polyn... 18 \\ = &2(z^2 - (3i)^2)\\ = &2(z + 3i)(z - 3i) \end{align} ====Question 1(iii)==== Factorize the pol... = & 3(z^2 - (11i)^2)\\ = & 3(z + 11i)(z - 11i) \end{align} ====Question 1(iv)==== Factorize the poly... {5}i\right)\left(z - \dfrac{\sqrt{3}}{5}i\right) \end{align} ====Question 1(v)==== Factorize the polyno
- Question 9, Exercise 1.2
- 2}{2^2+4^2} = \dfrac{2}{20}\\ &= \dfrac{1}{10}. \end{align} \begin{align} Im(2+4i)^{-1} & = Im(z^{-1}... 4}{2^2+4^2} = -\dfrac{4}{20}\\ &= \dfrac{1}{5}. \end{align} GOOD ====Question 9(ii)==== Find real and ... lign} |z| &= \sqrt{3^2 + (-2)^2} \\&= \sqrt{13}. \end{align} Then \[|z|^4=169.\] Using in above formul... 13})^4} \\ &= \frac{9 - 4}{169} = \frac{5}{169}. \end{align} \begin{align} Im((3 - 2i)^{-2}) &= \frac{-
- Question 1, Exercise 1.4
- rt{3})^2} \\ & = \sqrt{4 + 12} = \sqrt{16} = 4. \end{align} and \begin{align} \alpha & = \tan^{-1}\lef... ght|\\ & = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}. \end{align} Since the complex number \( 2 + i 2 \sqrt... })^2}\\ &= \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3}. \end{align} Next, \begin{align} \alpha &= \tan^{-1}\... t(\frac{1}{\sqrt{3}}\right) \\ &= \frac{\pi}{6}. \end{align} Since the complex number \( 3 - i \sqrt{3
- Question 7, Exercise 1.4
- -1 \\ \implies & y = -x+1 \\ \implies & x+y = 1. \end{align*} As required. =====Question 7(ii)===== Co... x^2+y^2 = 4\sqrt{1} \\ \implies & x^2+y^2 = 4. \end{align*} =====Question 7(iii)===== Convert the fol... es & -\sqrt{3}\leq \frac{y}{x-4} \leq \sqrt{3}. \end{align*} As required. =====Question 7(iv)===== Con... \sqrt{3}\left(\frac{y-x+4}{x+y-4}\right) \leq 1. \end{align*} As required. =====Question 7(v)===== Conv
- Question 4, Exercise 1.1
- 3i)x+(1+3i)y+2=0\\ \implies &(2x+y+2)+(3x+3y)i=0.\end{align} Comparing real and imaginary parts \begin{... +2&=0 \quad \cdots(1)\\ 3x+3y&=0\quad \cdots (2) \end{align} From (2), \begin{align} &3x=-3y \\ x=-y \quad ... (3) \end{align} Putting value of $x$ in (1) \begin{align}2(-y)+y+2&=0\\ -2y+y&=-2\\ -y&=-2\\ y&=2\end{align} Putting in $(3)$, we have $x=-2$. Hence $
- Question 8, Exercise 1.2
- (2y-1)|=4 \\ \implies & \sqrt{(2x)^2+(2y-1)^2}=4 \end{align} Squaring on both sides \begin{align} & (2x... +4y^2-4y+1-16=0 \\ \implies & 4x^2+4y^2-4y-15=0, \end{align} as required. GOOD ====Question 8(ii)==== W... mplies & \sqrt{(x-1)^2+y^2} = \sqrt{x^2+(y-1)^2} \end{align} Squaring both sides, we get \begin{align} ... -2y \\ \implies & x = y \\ \implies & x - y =0, \end{align} as required. GOOD ====Question 8(iii)====
- Question 2, Exercise 1.3
- s & (z - 3)^2+7= 0 \\ \implies & (z - 3)^2 = 7. \end{align} Take the square root of both sides: \begin... 3 = \pm \sqrt{7} \\ \implies &z = 3 \pm \sqrt{7}\end{align} Hence Solutioin set=$\{3 \pm \sqrt{7}\}$. ... \begin{align} -\dfrac{1}{2} z^{2} - 5z + 2& = 0 \end{align} Multiply through by $-2$ to eliminate the ... 0 \\ z^2 + 10z +25-25-4&=0\\ (z + 5)^2 - 29&=0 \end{align} Take the square root of both sides: \begin
- Question 6(i-ix), Exercise 1.4
- \sqrt{2}}-\dfrac{i}{\sqrt{2}} \right) \\ =& 1-i. \end{align} =====Question 6(ii)===== Write a given com... right) \\ =& -\frac{5\sqrt{3}}{2} - \frac{5}{2}i \end{align*} =====Question 6(iii)===== Write a given c... }\right) \\ =& 2\left(0 + i(-1)\right) \\ =& -2i \end{align*} =====Question 6(iv)===== Write a given co... i \cdot \frac{1}{2}\right) \\ =& -2\sqrt{3} + 2i \end{align*} =====Question 6(v)===== Write a given com
- Question 2, Exercise 1.1
- n{align}&(3+i2)+(2+i4)\\ =&(3+2)+(i2+i4)\\ =&5+i6\end{align} GOOD ====Question 2(ii)==== Write the fol... n{align}&(4+3i)-(2+5i)\\ =&(4-2)+(3i-5i)\\ =&2-2i\end{align} GOOD ====Question 2(iii)==== Write the fol... lign} &(4+7i)+(4-7i)\\ =&(4+4)+(7i-7i)\\ =&8+0i. \end{align} GOOD ====Question 2(iv)==== Write the fol... ign} &(2+5i)-(2-5i)\\ =&(2-2)+(5i+5i)\\ =&0+10i. \end{align} GOOD ====Question 2(v)==== Write the foll
- Question 7, Exercise 1.1
- n{align}|z|&= \sqrt{(11)^2+(12)^2}\\ &=\sqrt{265}\end{align} Hence $|11+12 i|=\sqrt{265}$. GOOD ====... 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{align} Hence $$|(2+3 i)-(2+6 i)|=3.$$ GOOD ====Q... qrt{5}\sqrt{45}\\ &=\sqrt{5}\cdot 3\sqrt{5} = 15.\end{align} Hence $|(2-i)(6+3 i)|=15$. GOOD ====Quest
- Question 3, Exercise 1.4
- \quad \theta=\tan^{-1}\left(\dfrac{b}{a}\right). \end{align*} We can write these complex numbers in pol... quad \text{and}\quad z=|z|e^{i\theta} \,\,-- (1) \end{align*} Now we have given \begin{align*} & \left(... mplies &z_1 \cdot z_2 \cdot z_3 \cdots z_n = z. \end{align*} By using $(1)$, we have \begin{align*} &|... _2+\theta_3+\ldots+\theta_n)} = |z| e^{i\theta}. \end{align*} This gives $$|z_1|\cdot|z_2|\cdot|z_3|\cd
- Question 1, Exercise 1.1
- 15}} \quad \because i^2=-1\\ &=i\cdot(-1)\\ &=-i.\end{align} GOOD ====Question 1(ii)==== Evaulate $... i \\ &=-(-1)^{11} \cdot i \\ &=-(-1)\cdot i = i. \end{align} GOOD ====Question 1(iii)==== Evaluate ${{\... \dfrac{i}{i}=\dfrac{i}{-\left( -1 \right)}\\ &=i\end{align} GOOD ====Question 1(iv)==== Evaluate $\d... \ &=i \cdot (i^2)^7 \\ &=i \cdot (-1)^7 \\ &= -i \end{align} GOOD ====Question 1(v)==== Evaluate $i^{2
- Question 3, Exercise 1.1
- =&\dfrac{7-9i}{2}\\ =&\dfrac{7}{2}-\dfrac{9}{2}i\end{align} GOOD ====Question 3(ii)==== Simplify the f... dfrac{7-i}{25}\\ =&\dfrac{7}{25}-\dfrac{1}{25}i. \end{align} GOOD ====Question 3(iii)==== Simplify the... {-2i}{9+1}\\ =&-\dfrac{2}{10}i\\ =&-\dfrac{1}{5}i\end{align} GOOD ====Question 3(iv)==== Simplify the f... rac{1}{-2i}\\ =&\dfrac{1}{2i}-\dfrac{1}{2i}\\ =&0\end{align} GOOD ====Question 3(v)==== Simplify: $(2+