Question 4, Exercise 1.3

Solutions of Question 4 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Solve the simultaneous system of linear equation with complex coefficients: $(1-i) z+(1+i) \omega=3 ; 2 z-(2+5 i) \omega=2+3 i$.

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(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}\times\dfrac{9-5i}{9-5i}\\ &=\dfrac{9-5-5i-9i}{81+25}\\ &=\dfrac{4-14i}{106}\\ &=\dfrac{2}{53}-\dfrac{7}{53}i\end{align} Put value of $\omega$ in (1), we have \begin{align} &(1-i) z+(1+i)\left(\dfrac{2}{53}-\dfrac{7}{53}i \right)=3\\ \implies &(1-i) z+\dfrac{2}{53}+\dfrac{7}{53}-\dfrac{7}{53}i+\dfrac{2}{53}i=3\\ \implies &(1-i) z+\dfrac{9}{53}-\dfrac{5}{53}i=3\\ \implies &(1-i) z=3-\dfrac{9}{53}-\dfrac{5}{53}i\\ \implies &(1-i) z=\dfrac{150}{53}-\dfrac{5}{53}i\\ \end{align} \begin{align} \implies z&=\dfrac{\dfrac{150}{53}-\dfrac{5}{53}i}{1-i}\\ &=\dfrac{1}{53}\dfrac{150-5i}{1-i}\\ &=\dfrac{1}{53}\dfrac{150-5i}{1-i}\times \dfrac{1+i}{1+i}\\ &=\dfrac{1}{53}\dfrac{150+5+150i-5i}{1+1}\\ &=\dfrac{1}{53}\dfrac{155+145i}{2}\\ &=\dfrac{155}{106}+\dfrac{145}{106}i\end{align} Thus, we have $$z=\dfrac{155}{106}+\dfrac{145}{106}i, \omega=\dfrac{2}{53}-\dfrac{7}{53}i.$$

Solve the simultaneous system of linear equation with complex coefficients: $2 i z+(3-2 i) \omega=1+i ;(1-2 i) z+(3+2 i) \omega=5+6 i$.

Solution.

\begin{align} &2 i z+(3-2 i) \omega=1+i \quad \cdots(1)\\ &(1-2 i) z+(3+2 i) \omega=5+6 i \quad \cdots(2) \end{align}

Multiplying $(1)$ by $(1-2i)$, we get:

\begin{align} &(1-2i)(2i z) + (1-2i)(3-2i) \omega = (1-2i)(1+i)\\ \implies &(2i+4)z + (3-4-2i-6i) \omega = 1+2+i-2i\\ \implies &(4+2i)z + (-1-8i) \omega = 3-i \quad \cdots(3) \end{align}

Multiplying equation (2) by $2i$, we get:

\begin{align} &2i(1-2i) z + 2i(3+2i) \omega = 2i(5+6i)\\ \implies &(2i+4) z + (6i-4) \omega = 10i-12 \\ \implies &(4+2i) z + (-4+6i) \omega = -12+10i \quad \cdots(4) \end{align} $(3)-(4)$,we have \begin{align} &(-1-8i+4-6i)\omega=3-i+12-10i\\ \implies &(3-14i)\omega=15-11i\\ \end{align} \begin{align} \implies \omega & =\dfrac{15-11i}{3-14i}\\ &= \dfrac{15 - 11i}{3 - 14i} \cdot \dfrac{3 + 14i}{3 + 14i}\\ &= \dfrac{45 + 154 + 210i - 33i}{9 + 196} \\ &= \dfrac{199 + 177i}{205}\\ \implies \omega & = \dfrac{199}{205} + \dfrac{177}{205}i. \end{align} Now, substituting $\omega$ back into equation $(1)$ to find $z$: \begin{align} &2iz + (3-2i)\left( \dfrac{199}{205} + \dfrac{177}{205}i\right) = 1 + i \\ \implies & 2iz + \dfrac{597}{205}+\dfrac{354}{205}-\dfrac{398}{205}i+\dfrac{531}{205}i= 1 + i\\ \implies & 2iz + \dfrac{951}{205}+\dfrac{133}{205}i = 1 + i\\ \implies & 2iz = 1 + i - \dfrac{951}{205}-\dfrac{133}{205}i\\ \implies & 2iz = -\dfrac{746}{205}+\dfrac{72}{205}i \end{align} Dividing by $2i$ \begin{align} z &= -\dfrac{373}{205i}+\dfrac{36i}{205i}\\ &= \dfrac{373}{205}i+\dfrac{36}{205} \quad \because \dfrac{1}{i}=-i\\ &=\dfrac{36}{205}+\dfrac{373}{205}i \end{align} Thus, we have: $$z = \dfrac{36}{205} + \dfrac{373}{205}i;\quad \omega = \dfrac{199}{205} + \dfrac{177}{205}i.$$

Solve the simultaneous system of linear equation with complex coefficients: $\dfrac{3}{i} z-(6+2 i) \omega=5 ; \quad \dfrac{i}{2} z+\left(\dfrac{3}{4}-\dfrac{1}{2} i\right) \omega=\left(\dfrac{1}{2}+2 i\right)$.

Solution. Given: \begin{align} &\dfrac{3}{i} z - (6 + 2i) \omega = 5\\ &-3i z - (6 + 2i) \omega = 5 \quad \cdots(1) \quad \because \dfrac{1}{i}=-i.\\ &\dfrac{i}{2} z + \left( \dfrac{3}{4} - \dfrac{1}{2}i\right) \omega = \left( \dfrac{1}{2} + 2i \right) \quad \cdots(2) \end{align} Multiply (1) by $2$ and (2) by $12$, we have \begin{align}-6i z - (12 + 4i) \omega = 10 \quad \cdots(3)\end{align} \begin{align} 6iz + (9-6i) \omega = 6+24i \quad \cdots(4) \end{align} $(3)+(4)$, we have \begin{align} &(-3-10i)\omega = 16+24i\\ \implies & \omega = \dfrac{16+24i}{-3-10i}\\ \,\,\, &= \dfrac{16+24i}{-3-10i}\times \dfrac{-3+10i}{-3+10i}\\ \,\,\, &= \dfrac{-288+88i}{109}\\ \implies &\omega= -\dfrac{288}{109}+ \dfrac{88}{109}i\end{align} Put vale of $\omega$ in $(1)$, we have \begin{align} & -3i z - (6 + 2i)\left( -\dfrac{288}{109} + \dfrac{88}{109}i\right) = 5\\ \implies &-3i z - \left(-\dfrac{1728}{109}- \dfrac{176}{109} + \dfrac{528}{109}i - \dfrac{576}{109}i \right) = 5\\ \implies &-3i z - \left(-\dfrac{1904}{109} - \dfrac{48}{109}i\right) = 5\\ \implies &-3i z = 5 - \dfrac{1904}{109} - \dfrac{48}{109}i\\ \implies &-3i z = -\dfrac{1359}{109} - \dfrac{48}{109}i\\ \implies & 3i z = \dfrac{1359}{109} + \dfrac{48}{109}i \end{align} Dividing by $3i$, we get \begin{align} &z = \dfrac{453}{109i} + \dfrac{16i}{109i}\\ \implies &z = \dfrac{16}{109}-\dfrac{453}{109}i \quad \because \frac{1}{i}=-i. \end{align}

Thus, we have: $$z = \dfrac{16}{109}-\dfrac{453}{109}i;\omega= -\dfrac{288}{109}+ \dfrac{88}{109}i$$

Solve the simultaneous system of linear equation with complex coefficients: $\dfrac{1}{1-i} z+(1+i) \omega=3 ; \quad \dfrac{2}{i} z-(2-3 i) \omega=2+6 i$.

Solution.

Given \begin{align} \dfrac{1}{1-i} z + (1+i) \omega &= 3 \quad \cdots(1)\\ \dfrac{2}{i} z - (2-3i) \omega &= 2 + 6i \quad \cdots(2) \end{align} Multiply (1) by $2(1-i)$, we have \begin{align} &2z+ 2(1+i)(1-i)\omega =6(1-i)\\ \implies &2z+4\omega=6-6i\quad\cdots(3)\end{align} Multiply $i$ by (2), we have \begin{align} 2z-(3+2i)\omega=-6+2i\quad \cdots(4) \end{align} $(3)-(4)$ we have, \begin{align} & 4\omega+(3+2i)\omega=-4i\\ \implies &(7+2i)\omega= 12-8i \end{align} \begin{align} \implies \omega&=\dfrac{12-8i}{7+2i}\\ &=\dfrac{12-8i}{7+2i}\times\dfrac{7-2i}{7-2i}\\ &=\dfrac{84-16-24i-56i^2}{49+4}\\ &=\dfrac{68-80i}{53}\\ &=\dfrac{68}{53}-\dfrac{80}{53}i\end{align} Putting value of $\omega$ in $(3)$, we get \begin{align} &2z+4\left(\dfrac{68}{53}-\dfrac{80}{53}i \right)=6-6i \\ \implies & 2z = 6-6i-\dfrac{272}{53}+\dfrac{320}{53}i\\ \implies & 2z = \dfrac{46}{53}+\dfrac{2}{53}i\\ \implies & z = \dfrac{23}{53}+\dfrac{1}{53}i \end{align} Hence, we have $$z = \dfrac{23}{53}+\dfrac{1}{53}i; \omega=\dfrac{68}{53}-\dfrac{80}{53}i.$$

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