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- Question 4, Exercise 1.3
- 4(i)===== Solve the simultaneous system of linear equation with complex coefficients: $(1-i) z+(1+i) \omega=... (ii)===== Solve the simultaneous system of linear equation with complex coefficients: $2 i z+(3-2 i) \omega=... ga = 3-i \quad \cdots(3) \end{align} Multiplying equation (2) by $2i$, we get: \begin{align} &2i(1-2i) z +... \end{align} Now, substituting $\omega$ back into equation $(1)$ to find $z$: \begin{align} &2iz + (3-2i)\le
- Question 2, Exercise 1.3
- abad, Pakistan. ====Question 2(i)==== Solve the equation by completing square: $z^{2}-6 z+2=0$. **Solutio... m \sqrt{7}\}$. ====Question 2(ii)==== Solve the equation by completing square: $-\dfrac{1}{2} z^{2}-5 z+2=... \sqrt{29}\}$ ====Question 2(iii)==== Solve the equation by completing square: $4 z^{2}+5 z=14$. **Soluti... 9}}{8}\right\}$ ====Question 2(iv)==== Solve the equation by completing square: $z^{2}=5 z-3$. **Solution.
- Question 3, Exercise 1.3
- stan. ====Question 3(i)==== Solve the quadratic equation: $\dfrac{1}{3} z^{2}+2 z-16=0$. **Solution.** Gi... \}$. ====Question 3(ii)==== Solve the quadratic equation: $z^{2}-\frac{1}{2} z+17=0$. **Solution.** Give... \} $ ====Question 3(iii)==== Solve the quadratic equation: $z^{2}-6 z+25=0$. **Solution.** Given $$ z^{2}... 4i\}$ ====Question 3(iv)==== Solve the quadratic equation: $z^{2}-9 z+11=0$. **Solution.** Given $$z^{2}
- Question 7, Exercise 1.4
- . =====Question 7(i)===== Convert the following equation in Cartesian form: $\arg (z-1)=-\dfrac{\pi}{4}$