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- Question 6(i-ix), Exercise 1.4
- bad, Pakistan. =====Question 6(i)===== Write a given complex number in the algebraic form: $\sqrt{2}\l... 1-i. \end{align} =====Question 6(ii)===== Write a given complex number in the algebraic form: $5\left(\co... }i \end{align*} =====Question 6(iii)===== Write a given complex number in the algebraic form: $2\left(\co... -2i \end{align*} =====Question 6(iv)===== Write a given complex number in the algebraic form: $4\left(\co
- Question 6(x-xvii), Exercise 1.4
- abad, Pakistan. =====Question 6(x)===== Write a given complex number in the algebraic form: $7 \sqrt{2}... revious parts.// =====Question 6(xi)===== Write a given complex number in the algebraic form: $10 \sqrt{2... vious parts.// =====Question 6(xii)===== Write a given complex number in the algebraic form: $2\left(\co... vious parts.// =====Question 6(xiii)===== Write a given complex number in the algebraic form: $\dfrac{1}{
- Question 8, Exercise 1.2
- $x$ and $y$ by taking $z=x+i y$. **Solution.** Given: $$|2z-i|=4.$$ Put $z=x+i y$, we have \begin{alig... $x$ and $y$ by taking $z=x+i y$. **Solution.** Given: $$|z-1|=|\bar{z}+i|.$$ Put $z=x+iy$, we have \be... $x$ and $y$ by taking $z=x+i y$. **Solution.** Given: $$|z-4i| + |z+4i| = 10.$$ Put $z = x + iy$, we h... $x$ and $y$ by taking $z=x+i y$. **Solution.** Given: $$\dfrac{1}{2} Re(i \bar{z}) = 4.$$ Put $z = x
- Question 10, Exercise 1.2
- line{z_{1}}}{\overline{z_{2}}}$. **Solution.** Given \[z_1 = -3 + 2i, \quad z_2 = 1 - 3i\] Then \begin... ne{z_{1}}\,\, \overline{z_{2}}$. **Solution.** Given \[ z_1 = -3 + 2i, \quad z_2 = 1 - 3i. \] First w... erline{z_{1}}+\overline{z_{2}}$. **Solution.** Given \[ z_1 = -3 + 2i, \quad z_2 = 1 - 3i. \] First w... z_{1}\right|\left|z_{2}\right|$. **Solution.** Given: \begin{align} z_1 = -3 + 2i, \quad z_2 = 1 - 3i
- Question 3, Exercise 1.3
- on: $\dfrac{1}{3} z^{2}+2 z-16=0$. **Solution.** Given \begin{align}&\dfrac{1}{3}z^{2}+2 z-16=0\\ \impli... tion: $z^{2}-\frac{1}{2} z+17=0$. **Solution.** Given $$ z^{2} - \frac{1}{2}z + 17 = 0 $$ Using the qua... ratic equation: $z^{2}-6 z+25=0$. **Solution.** Given $$ z^{2} - 6z + 25 = 0 $$ Using the quadratic for... ratic equation: $z^{2}-9 z+11=0$. **Solution.** Given $$z^{2} - 9z + 11 = 0 $$ Using the quadratic form
- Question 6, Exercise 1.1
- te of the complex number $4-3 i$. **Solution.** Given: $z=4-3 i$, then $\bar{z}=4+3i$. ====Question 6... number $2+\sqrt{\dfrac{-1}{5}}$. **Solution.** Given: \begin{align}z=&2+\sqrt{\dfrac{-1}{5}}\\ =&2+\sq... r $\dfrac{5 }{2}i-\dfrac{7}{8}$. **Solution.** Given: $z=\dfrac{5 }{2}i-\dfrac{7}{8},$ then $\bar{z}=-
- Question 9, Exercise 1.2
- eft(x_{1}^{2}+y_{1}^{2}\right)^{2}}.\end{align} Given \(z_1 = 4 + 2i\) and \(z_2 = 2 + 5i\), we have: ... {3-7 i}{2+5 i}$. **Solution.** We will use the given formulas: \[\text{Re}\left(\frac{x_{1}+i y_{1}}... _{2} y_{1}-x_{1} y_{2}}{x_{2}^{2}+y_{2}^{2}}.\] Given \(z_1 = 3 - 7i\) and \(z_2 = 2 + 5i\), we have: \
- Question 4, Exercise 1.3
- =\left(\dfrac{1}{2}+2 i\right)$. ** Solution. ** Given: \begin{align} &\dfrac{3}{i} z - (6 + 2i) \omega ... {2}{i} z-(2-3 i) \omega=2+6 i$. ** Solution. ** Given \begin{align} \dfrac{1}{1-i} z + (1+i) \omega &=
- Question 3, Exercise 1.4
- z|e^{i\theta} \,\,-- (1) \end{align*} Now we have given \begin{align*} & \left(x_{1}+i y_{1}\right)\left(... ts. **Alternative Method for Part (i)** We have given \begin{align*} & \left(x_{1}+i y_{1}\right)\left(
- Question 9, Exercise 1.4
- qrt{221}}{17} \neq 1.$$ <fc #ff0000>The contents, given in the textbook, related to these question are no... }=2$. ** Solution. ** <fc #ff0000>The contents, given in the textbook, related to these question are no
- Question 4, Exercise 1.2
- |=16$ find $\left|z_{2}\right|$. **Solution.** Given $$z_{1}=2-3i$$ Then \begin{align}|z_1|&=\sqrt{(2)
- Question 6, Exercise 1.2
- ere $z_{1}=3+i$ and $z_{2}=1+i$. **Solution.** Given: \begin{align} &z_{1}=3+i\text{ and } z_{2}=1+i.\
- Question 1, Exercise 1.3
- nctions: $z^{3}+2 z^{2}-23 z-60$. **Solution.** Given: $$z^{3}+2 z^{2}-23 z-60$$ Putting $z = -3$: \be
- Question 4, Exercise 1.4
- a$, show that $z=i \tan \theta$ ** Solution. ** Given \begin{align}&\dfrac{1+z}{1-z}=\cos 2 \theta+i \
- Question 5, Exercise 1.4
- =3 \sin (\alpha+\beta+\gamma)$. ** Solution. ** Given: \begin{align} \cos \alpha + \cos \beta + \cos \g