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- Question 5, Exercise 1.3
- }+z+3=0$\\ According to the quadratic formula, we have\\ $a=1,\,\,\,b=1$ and $c=3$\\ Quadratic formula ... }-z-1=0$\\ According to the quadratic formula, we have\\ $a=1,\quad b=-1$ and $c=-1$\\ Quadratic formul... -2z+i=0$\\ According to the quadratic formula, we have\\ $a=1,\,\,\,b=-2$ and $c=i$\\ Quadratic formula... 2}}+4=0$\\ According to the quadratic formula, we have\\ $a=1,\,\,\,b=0$ and $c=4$\\ Quadratic formula i
- Question 5, Exercise 1.2
- &=3-i \ldots (2)\end{align} From (1) and (2), we have $$\overline{z_1+z_2}=\overline{z_1}+\overline{z_... \ &=13 \ldots (2)\end{align} From (1) and (2), we have $$\overline{z_1 z_2}=\overline{z_1}\overline{z_2}... {2}}} \ldots (2) \end{align} From (1) and (2), we have $$\overline{\left(\dfrac{z_1}{z_2}\right)}=\dfrac
- Question 6, Exercise 1.3
- ight)=0$\\ According to the quadratic formula, we have\\ $a=1,\,\,\,b=-2$ and $c=4$ \\ Quadratic formula... -3z+3=0$\\ According to the quadratic formula, we have\\ $a=1,\,\,\,b=-3$ and $c=3$ \\ Quadratic formula... end{align} According to the quadratic formula, we have\\ $a=1,\quad\quad b=1$ and $c=1$ \\ Quadratic for
- Question 11, Exercise 1.1
- dfrac{-2}{5}+\dfrac{11i}{5} \end{align} Hence, we have $${\rm Re}\left( \dfrac{{{z}_{1}}{{z}_{2}}}{\over... c{1}{z_1\overline{z_1}}=\dfrac{1}{5}.$$ Hence, we have $${\rm Im}\left(\dfrac{1}{z_1\overline{z_1}}\righ
- Question 2, Exercise 1.3
- z}^{3}}+6z+20$.\\ By using synthetic division, we have $$\begin{array}{c|cccc} -2 & 1 & 0 & 6 & 20 \\ ... {{z}^{2}}+z-2$.\\ By using synthetic division, we have $$\begin{array}{c|cccc} 2 & 1 & -2 & 1 & -2 \\
- Question 2, Exercise 1.2
- =4-3i \ldots (2) \end{align} From (1) and (2), we have the required result. Now, we prove associative p
- Question 6, Exercise 1.2
- native Method**\\ We know $|z|^2=z\bar{z}$, so we have \begin{align}|{{z}_{1}}{{z}_{2}}{{|}^{2}}&={{z}_{
- Question 8, Exercise 1.2
- $z=a+bi$ ... (1) Then $\overline{z}=a-bi$. We have given \begin{align}&z=\overline{z} \\ \implies &a
- Question 3 & 4, Exercise 1.3
- -2z+5=0$\\ According to the quadratic formula, we have\\ $a=1,\quad b=-2$ and $c=5$\\ Quadratic formula
- Question 6, 7 & 8, Review Exercise 1
- end{align} According to the quadratic formula, we have\\ $$a=1,\quad b=-2$$ and $$c=2$$\\ Quadratic for