4. Solve the simultaneous system of linear equations with complex coe (i) \( (1-i) z+(1+i) \omega=3 ; \quad 2 z-(2+5 i) \omega=2+3 i \) (ii) \( 2 i z+(3-2 i) \omega=1+i ; \quad(1-2 i) z+(3+2 i) \omega=5+6 i \) (iii) \( \frac{3}{i} z-(6+2 i) \omega=5 ; \quad \frac{i}{2} z+\left(\frac{3}{4}-\frac{1}{2} i\right) \omega=\left(\frac{1}{2}+2 i\right) \) (iv) \( \frac{1}{1-i} z+(1+i) \omega=3 ; \quad \frac{2}{i} z-(2-3 i) \omega=2+6 i \)

Solve the system of equations \( (1-i)z+(1+i)\omega=3;2z-(2+5i)\omega=2+3i \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}\left(1-i\right)z+\left(1+i\right)\omega =3\\2z-\left(2+5i\right)\omega =2+3i\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}\left(1-i\right)z+\left(1+i\right)\omega =3\\2z+\left(-2-5i\right)\omega =2+3i\end{array}\right.\) - step2: Solve the equation: \(\left\{ \begin{array}{l}\omega =\frac{3}{2}\\2z+\left(-2-5i\right)\omega =2+3i\end{array}\right.\) - step3: Substitute the value of \(\omega:\) \(2z+\left(-2-5i\right)\times \frac{3}{2}=2+3i\) - step4: Multiply the numbers: \(2z-3-\frac{15}{2}i=2+3i\) - step5: Move the constant to the right side: \(2z=2+3i+3+\frac{15}{2}i\) - step6: Add the numbers: \(2z=5+\frac{21}{2}i\) - step7: Divide both sides: \(\frac{2z}{2}=\frac{5+\frac{21}{2}i}{2}\) - step8: Divide the numbers: \(z=\frac{5}{2}+\frac{21}{4}i\) - step9: Calculate: \(\left\{ \begin{array}{l}\omega \notin \mathbb{R}\\z \notin \mathbb{R}\end{array}\right.\) - step10: Calculate: \(\left(\omega,z\right) \notin \mathbb{R}^{2}\) - step11: Alternative Form: \(\textrm{No real solution}\) Solve the system of equations \( 2iz+(3-2i)\omega=1+i;(1-2i)z+(3+2i)\omega=5+6i \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}2iz+\left(3-2i\right)\omega =1+i\\\left(1-2i\right)z+\left(3+2i\right)\omega =5+6i\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}\omega =\frac{1}{3}\\\left(1-2i\right)z+\left(3+2i\right)\omega =5+6i\end{array}\right.\) - step2: Substitute the value of \(\omega:\) \(\left(1-2i\right)z+\left(3+2i\right)\times \frac{1}{3}=5+6i\) - step3: Multiply the numbers: \(\left(1-2i\right)z+1+\frac{2}{3}i=5+6i\) - step4: Move the constant to the right side: \(\left(1-2i\right)z=5+6i-\left(1+\frac{2}{3}i\right)\) - step5: Subtract the terms: \(\left(1-2i\right)z=4+\frac{16}{3}i\) - step6: Divide both sides: \(\frac{\left(1-2i\right)z}{1-2i}=\frac{4+\frac{16}{3}i}{1-2i}\) - step7: Divide the numbers: \(z=-\frac{4}{3}+\frac{8}{3}i\) - step8: Calculate: \(\left\{ \begin{array}{l}\omega \notin \mathbb{R}\\z \notin \mathbb{R}\end{array}\right.\) - step9: Calculate: \(\left(\omega,z\right) \notin \mathbb{R}^{2}\) - step10: Alternative Form: \(\textrm{No real solution}\) Solve the system of equations \( \frac{3}{i}z-(6+2i)\omega=5;\frac{i}{2}z+\left(\frac{3}{4}-\frac{1}{2}i\right)\omega=\left(\frac{1}{2}+2i\right) \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}\frac{3}{i}\times z-\left(6+2i\right)\omega =5\\\frac{i}{2}z+\left(\frac{3}{4}-\frac{1}{2}i\right)\omega =\frac{1}{2}+2i\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}-3iz+\left(-6-2i\right)\omega =5\\\frac{1}{2}iz+\left(\frac{3}{4}-\frac{1}{2}i\right)\omega =\frac{1}{2}+2i\end{array}\right.\) - step2: Solve the equation: \(\left\{ \begin{array}{l}\omega =-\frac{5}{6}\\\frac{1}{2}iz+\left(\frac{3}{4}-\frac{1}{2}i\right)\omega =\frac{1}{2}+2i\end{array}\right.\) - step3: Substitute the value of \(\omega:\) \(\frac{1}{2}iz+\left(\frac{3}{4}-\frac{1}{2}i\right)\left(-\frac{5}{6}\right)=\frac{1}{2}+2i\) - step4: Multiply the numbers: \(\frac{1}{2}iz-\frac{5}{8}+\frac{5}{12}i=\frac{1}{2}+2i\) - step5: Move the constant to the right side: \(\frac{1}{2}iz=\frac{1}{2}+2i+\frac{5}{8}-\frac{5}{12}i\) - step6: Add the numbers: \(\frac{1}{2}iz=\frac{9}{8}+\frac{19}{12}i\) - step7: Divide both sides: \(\frac{\frac{1}{2}iz}{\frac{1}{2}i}=\frac{\frac{9}{8}+\frac{19}{12}i}{\frac{1}{2}i}\) - step8: Divide the numbers: \(z=\frac{19}{6}-\frac{9}{4}i\) - step9: Calculate: \(\left\{ \begin{array}{l}\omega \notin \mathbb{R}\\z \notin \mathbb{R}\end{array}\right.\) - step10: Calculate: \(\left(\omega,z\right) \notin \mathbb{R}^{2}\) - step11: Alternative Form: \(\textrm{No real solution}\) Solve the system of equations \( \frac{1}{1-i}z+(1+i)\omega=3;\frac{2}{i}z-(2-3i)\omega=2+6i \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}\frac{1}{1-i}\times z+\left(1+i\right)\omega =3\\\frac{2}{i}\times z-\left(2-3i\right)\omega =2+6i\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}\left(\frac{1}{2}+\frac{1}{2}i\right)z+\left(1+i\right)\omega =3\\-2iz+\left(-2+3i\right)\omega =2+6i\end{array}\right.\) - step2: Calculate: \(\left\{ \begin{array}{l}\omega \in \varnothing \\z \in \varnothing \end{array}\right.\) - step3: Rewrite: \((\omega, z) \in \varnothing\) The simultaneous system of linear equations with complex coefficients does not have a real solution for any of the given systems.