Search
You can find the results of your search below.
Fulltext results:
- Question 7, Exercise 1.4
- as \begin{align*} &\arg (z-1)=-\dfrac{\pi}{4} \\ \implies & \arg(x+iy-1) = -\dfrac{\pi}{4} \\ \implies & \arg(x-1+iy) = -\dfrac{\pi}{4} \\ \implies & \tan^{-1}\left(\dfrac{y}{x-1}\right) = -\dfrac{\pi}{4} \\ \implies & \dfrac{y}{x-1} = \tan\left(-\dfrac{\pi}{4}\righ
- Question 8, Exercise 1.2
- 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)===
- Question 4, Exercise 1.3
- ign} &2(1-i)z-(1-i) (2+5 i)\omega=(1-i) (2+3i)\\ \implies & (2-2i)z-(2+5+5i-2i)\omega=2+3+3i-2i \\ \implies & (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-i}{9+5i}\\ &=\dfrac{1-i}{9+5i}
- Question 4, Exercise 1.1
- olution.** \begin{align}&(2+3i)x+(1+3i)y+2=0\\ \implies &(2x+y+2)+(3x+3y)i=0.\end{align} Comparing real a... egin{align}&\dfrac{x}{(1+i)}+\dfrac{y}{1-2i}=1\\ \implies &\dfrac{x(1-2i)+y(1+i)}{(1+i)(1-2i)}=1\\ \implies &\dfrac{x-i2x+y+iy}{1-2i^2+i-2i}=1\\ \implies &\dfrac{(x+y)+(y-2x)i}{3-i}=1\\ \implies & (x+y)+(y-2x)i=3-i
- Question 4, Review Exercise
- gn*} & \left|\dfrac{z + 2i}{z - 2i}\right| = 1\\ \implies & |z + 2i| = |z - 2i|\\ \implies & |x + i(y + 2)| = |x + i(y - 2)|\\ \implies & \sqrt{x^2 + (y + 2)^2} = \sqrt{x^2 + (y - 2)^2} \end{align... in{align*} & x^2 + (y + 2)^2 = x^2 + (y - 2)^2\\ \implies & (y + 2)^2 = (y - 2)^2\\ \implies & y^2 + 4y +
- Question 3, Exercise 1.4
- \right)\ldots \left(x_{n}+i y_{n}\right)=a+i b\\ \implies &z_1 \cdot z_2 \cdot z_3 \cdots z_n = z. \end{al... 3} \cdots |z_n| e^{i\theta_n}=|z| e^{i\theta} \\ \implies & \left(|z_1|\cdot|z_2|\cdot|z_3|\cdots |z_n|\rig... 2\cdot|z_2|^2\cdot|z_3|^2\cdots |z_n|^2 = |z|^2$$ implies $$\left(x_{1}^{2}+y_{1}^{2}\right)\left(x_{2}^{2}... + 2k\pi, \quad \text{where } k\in \mathbb{Z}, $$ implies $$\sum_{r=1}^{n} \tan ^{-1}\left(\frac{y_{r}}{x_{
- Question 7, Review Exercise
- olution. ** \begin{align*} &2 z^{2}-11 z+16=0\\ \implies&z^2 - \dfrac{11}{2}z + 8 = 0\\ \implies& z^2 - \dfrac{11}{2}z = -8\\ \implies& z^2 - 2z\dfrac{11}{4}z + \dfrac{121}{16} = -8 + \dfrac{121}{16}\\ \implies&\left(z-\dfrac{11}{4}\right)^2 = -\dfrac{128}{16
- Question 10, Exercise 1.2
- \begin{align} -z_1 &= -(-3 + 2i) = 3 - 2i\\ \implies |-z_1| &= \sqrt{(3)^2 + (-2)^2} \\ &= \sqrt{9 + 4... Also \begin{align} \overline{z_1} &= -3 - 2i \\ \implies |\overline{z_1}| &= \sqrt{(-3)^2 + (-2)^2} \\ & ... align} -\overline{z_1} &= -(-3 - 2i) = 3 + 2i \\ \implies |-\overline{z_1}| &= \sqrt{(3)^2 + (2)^2} \\ & = ... 9(-1)} \\ &= \frac{-9 + 7i}{10} \end{align} $$\implies \frac{\overline{z_1}}{\overline{z_2}} = -\frac{9}
- Question 5, Review Exercise
- ion. ** \begin{align*} &(z-3 i)(2+5 i)=3-4 i \\ \implies & z-3 i=\dfrac{3-4 i}{2+5 i} \\ \implies & z-3 i=\dfrac{(3-4 i)(2-5i)}{(2+5 i)(2-5i)}\\ \implies & z-3 i=\dfrac{6-20-15i-8i}{4+25}\\ \implies & z-3 i=\dfrac{-14-23i}{29}\\ \implies & z=-\dfrac{14}{29}-\d
- Question 6, Exercise 1.2
- {z_{1}}{z_{2}}+\lambda\right|=\sqrt{\lambda+2}\\ \implies & |2-i+\lambda|=\sqrt{\lambda+2} \\ \implies & |2+\lambda-i|^2=\lambda+2 \\ \implies &(2+\lambda)^2+1=\lambda+2 \\ \implies &4+4\lambda+\lambda^2+1-\lambda-2=0 \\ \implies &\lambda^2+3\lambda
- Question 7, Exercise 1.2
- egin{align} &\left(|x|-|y|)^{2} \geq 0\right) \\ \implies & |x|^2+|y|^2-2|x||y| \geq 0 \\ \implies & |x|^2+|y|^2 \geq 2|x||y| \\ \implies & 2|x|^2+2|y|^2 \geq |x|^2+|y|^2+2|x||y| \\ \implies & 2(x^2+y^2) \geq\left(|x|+|y|\right)^2 \quad \becaus
- Question 2, Exercise 1.3
- Solution.** \begin{align} & z^2 - 6z + 2 = 0 \\ \implies & z^2 - 2(3)(z)+9-9+2=0 \\ \implies & (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 se
- Question 8, Review Exercise
- formula \begin{align} &x=x_{\max} e^{i\theta} \\ \implies & \sqrt{2} + i \sqrt{2}=x_{\max} e^{i\dfrac{\pi}{4}} \\ \implies & x_{\max} \left(\cos\dfrac{\pi}{4} +i\sin\dfrac{\pi}{4}\right)=\sqrt{2} + i \sqrt{2} \\ \implies & x_{\max} \left(\dfrac{1}{\sqrt{2}} +i\dfrac{1}{\sqrt{2}}\right)=\sqrt{2} + i \sqrt{2} \\ \implies & \dfrac{x_{\max}}{2} \left(\sqrt{2} +i\sqrt{2}\r
- Question 5, Exercise 1.1
- e \begin{align}&4z-3\bar{z}=\dfrac{1-18i}{2-i}\\ \implies &4(x+iy)-3(x-iy)=\dfrac{1-18i}{2-i}\times \dfrac{2+i}{2+i}\\ \implies &4x+4iy-3x+3iy=\dfrac{(1-18i)(2+i)}{2^2-i^2} \end{align} \begin{align} \implies x+7iy&=\dfrac{2-18i^2-36i+i}{4+1}\\ &=\dfrac{20-35i}{5}\\ &=\dfrac{5(4-7i)}{5}\\ \implies x+7iy&=4-7i\end{align} Equating real and imaginar
- Question 4, Exercise 1.4
- \dfrac{1+z}{1-z}=\cos 2 \theta+i \sin 2 \theta\\ \implies &\dfrac{1+z}{1-z}=e^{i2\theta}\\ \implies &(1+z)=(1-z)e^{i2\theta}\\ \implies &z+z e^{i2\theta}=e^{i2\theta}-1\ \end{align} \begin{align} z & =\dfra... cos\theta +i\sin\theta}\right)\\ \end{align} This implies $$ z=i\tan \theta. $$ ====Go to ==== <text align=