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- Question 4, Exercise 1.3
- i}\\ &=\dfrac{9-5-5i-9i}{81+25}\\ &=\dfrac{4-14i}{106}\\ &=\dfrac{2}{53}-\dfrac{7}{53}i\end{align} Put... =\dfrac{1}{53}\dfrac{155+145i}{2}\\ &=\dfrac{155}{106}+\dfrac{145}{106}i\end{align} Thus, we have $$z=\dfrac{155}{106}+\dfrac{145}{106}i, \omega=\dfrac{2}{53}-\dfrac{7}{53
- Question 10, Exercise 1.2
- ====== Question 10, Exercise 1.2 ====== Solutions of Question 10 of Exercise 1.2 of Unit 01: Complex Numbers. This is un... extbook Board, Islamabad, Pakistan. ====Question 10(i)==== For $z_{1}=-3+2 i$, verify: $$\left|z_{1}\... ht| = \sqrt{13}.$$ As required. GOOD ====Question 10(ii)==== For $z_{1}=-3+2 i$ and $z_{2}=1-3 i$ veri
- Question 10, Exercise 1.4
- ====== Question 10, Exercise 1.4 ====== Solutions of Question 10 of Exercise 1.4 of Unit 01: Complex Numbers. This is un... tbook Board, Islamabad, Pakistan. =====Question 10(i)===== Find the impedance $Z$ for the following values: $E=(-50+100 i)$ volts, $I=(-6-2 i)$ amps. ** Solution. **
- Question 8, Exercise 1.2
- OD ====Question 8(iii)==== Write $|z-4 i|+|z+4 i|=10$ in terms of $x$ and $y$ by taking $z=x+i y$. **Solution.** Given: $$|z-4i| + |z+4i| = 10.$$ Put $z = x + iy$, we have \begin{align} & |(x + iy) - 4i| + |(x + iy) + 4i| = 10 \\ \implies & |x + i(y - 4)| + |x + i(y + 4)| = 10 \\ \implies & \sqrt{x^2 + (y - 4)^2} + \sqrt{x^2 +
- Question 4, Exercise 1.1
- ightarrow \quad & \dfrac{x}{(2+i)}=\dfrac{(2+5i^2-10i-i)+(3y-2yi)}{(6+2i^2-4i-3i)}\\ \Rightarrow \quad... y$ in each of the following: $x(1+i)^2+y(2-i)^2=3+10i$ **Solution.** \begin{align}&x(1+i)^2+y(2-i)^2=3+10i\\ \Rightarrow \quad &x(1+i^2+2i)+y(4+i^2-4i)=3+10i\\ \Rightarrow \quad &x(1-1+2i)+y(4-1-4i)=3+10i\\
- Question 1, Exercise 1.3
- olynomial into linear functions: $2 z^{3}+3 z^{2}-10 z-15$. **Solution.** Suppose $$ P(z) = 2 z^{3}+3 z^{2}-10 z-15.$$ Since \begin{align}P\left(-\dfrac{3}{2} \... {3}{2}\right)^3 + 3\left(-\dfrac{3}{2}\right)^2 - 10\left(-\dfrac{3}{2}\right) - 15\\ &= -\dfrac{27}{4... n} \begin{array}{r|rrrr} -\dfrac{3}{2} & 2 & 3 & -10 & -15 \\ & & -3 & 0 & 15 \\ \hline & 2 & 0 & -1
- Question 3, Review Exercise
- tion 3(i) ===== Factorize the following: $3 x^{2}+108$ ** Solution. ** \begin{align*} & 3 x^{2}+108\\ =&3 (x^{2}+36)\\ =&3 (x^{2}-(6i)^2)\\ =&3 (x+6i)(x... tion. ** \begin{align*} &4 x^{2}+40\\ =&4 (x^{2}+10)\\ =&4 (x^{2}+(\sqrt{10}i)^2)\\ =&4 (x+\sqrt{10}i)(x-\sqrt{10}i) \end{align*} ====Go to ==== <te
- Question 2, Exercise 1.1
- gin{align} &(2+5i)-(2-5i)\\ =&(2-2)+(5i+5i)\\ =&0+10i. \end{align} GOOD ====Question 2(v)==== Write t... \\ =&\dfrac{9+2i^2+6i+3i}{9+1}\\ =&\dfrac{9-2+9i}{10}\\ =&\dfrac{7+9i}{10}\\ =&\dfrac{7}{10}+\dfrac{9}{10}i\end{align} GOOD ====Question 2(vii)==== Write the following comple
- Question 8, Exercise 1.4
- eta} \\ &=0.004 e^{i\dfrac{\pi}{4}} \\ &=\frac{4}{1000} \left(\cos\left(\dfrac{\pi}{4}\right) +i \sin\... } \\ &= 0.004 e^{i\dfrac{\pi}{3}} \\ &= \dfrac{4}{1000} (\cos(\dfrac{\pi}{3}) + i \sin(\dfrac{\pi}{3})) \\ &= \dfrac{4}{1000}(\dfrac{1}{2} + i \dfrac{\sqrt{3}}{2}) \\ &= \d... } \\ &= 0.004 e^{i\dfrac{\pi}{6}} \\ &= \dfrac{4}{1000} (\cos(\dfrac{\pi}{6}) + i \sin(\dfrac{\pi}{6})
- Question 1, Review Exercise
- alue of $(\sqrt{-25})(\sqrt{-4})$ is * (a) $10$ * (b) $-10$ * %%(c)%% $10 i$ * (d) $-10 i$ \\ <btn type="link" collapse="a9">See Answer</btn><collapse id="a9" collapsed="t
- Question 9, Exercise 1.2
- dfrac{2}{2^2+4^2} = \dfrac{2}{20}\\ &= \dfrac{1}{10}. \end{align} \begin{align} Im(2+4i)^{-1} & = Im... {20^2}\\ = &\frac{-252 + 320}{400}\\ = &\frac{17}{100}. \end{align} Next, we compute the imaginary p... (\dfrac{4+2i}{2+5i}\right)^{-2}\) is \(\dfrac{17}{100}\) and the imaginary part is \(\dfrac{36}{25}\).... text align="right"><btn type="success">[[math-11-nbf:sol:unit01:ex1-2-p10|Question 10 >]]</btn></text>
- Question 6(i-ix), Exercise 1.4
- mplex number in the algebraic form: $5\left(\cos 210^{\circ}+i \sin 210^{\circ}\right)$ ** Solution. ** \begin{align*} &5\left(\cos 210^\circ + i \sin 210^\circ\right) \\ =& 5\left(-\frac{\sqrt{3}}{2} - \frac{1}{2}i\right) \\ =& -\frac{5
- Question 2, Review Exercise
- ue of the following: $i^{2}+i^{4}+i^{6}+\cdots+i^{100}$\\ ** Solution. ** \begin{align*} & i^{2}+i^{4}+i^{6}+\ldots+i^{100} \\ =& i^2 + (i^2)^2 + (i^2)^3 + (i^2)^4 + \ldot... rac{(2-3 i)(3-5i)}{(3+5 i)(3-5i)}\\ =&\dfrac{6-15-10i-9i}{9+25}\\ =&\dfrac{-9-19i}{34}\\ =&-\dfrac{9}{
- Question 1, Exercise 1.1
- }\cdot i \\ =&(i^2)^{11}\cdot i+(i^2)^{29}+(i^2)^{10}\cdot i \\ =&(-1)^{11}\cdot i+(-1)^{29}+(-1)^{10}\cdot i \\ =& -i - i +i = -i \end{align} GOOD ====Go
- Question 2, Exercise 1.3
- 2$ to eliminate the fraction: \begin{align} z^2 + 10z - 4 &= 0 \\ z^2 + 10z +25-25-4&=0\\ (z + 5)^2 - 29&=0 \end{align} Take the square root of both sid