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- Question 2, Exercise 1.3
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... == Given: $$p\left( z \right)={{z}^{3}}+6z+20$$ By factor theorem, $(z-a)$ is a factor of $P(z)$ iff... } Thus $z+2$ is a factor of ${{z}^{3}}+6z+20$.\\ By using synthetic division, we have $$\begin{array}... $$P\left( z \right)={{z}^{3}}-2{{z}^{2}}+z-2$$ \\ By factor theorem $(z-a)$ is a factor of $()$ iff $P
- Question 1, Exercise 1.3
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... …(i)\\ 2z+3w&=11-5i …(ii)\end{align} Multiply $2$ by (i), we get\\ \begin{align}2z-8w&=6i …(iii)\end{... i …(i)\\ 2z+3w&=2 …(ii)\end{align} Multiply $2$ by (i), we get\\ \begin{align}2z+2w&=6i …(iii)\end{... i …(ii)\end{align} Multiply $\left( 2-i \right)$ by (ii), we get,\\ \begin{align}\left( 2-i \right)\l
- Question 7, Exercise 1.2
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... {2+3i}{5-2i}\times \dfrac{5+2i}{5+2i} \quad \text{by rationalizing} \\ =&\dfrac{10-6+15i+4i}{25+4}\\ =... -3+4i}{1-3i}\times \dfrac{1+3i}{1+3i} \quad \text{by rationalizing}\\ =&\dfrac{-3-12+4i-9i}{1+9}\\ =&
- Question 6, Exercise 1.3
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... 0$$ $$( z^2+z+1 )( z^2-z+1 )=0$$ $$(z^2+z+1 )=0$$ By using quadratic formula, we have $$z=\dfrac{-1\pm... $$z=\dfrac{-1\pm \sqrt{3}i}{2}$$ $$(z^2-z+1 )=0$$ By using quadratic formula, we have $$z=\dfrac{1\pm
- Question 5, Exercise 1.2
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... 2a-3bi}\times \dfrac{2a+3bi}{2a+3bi} \quad (\text{by rationalizing})\\ &=\dfrac{-2{{a}^{2}}+9{{b}^{2}}
- Question 6, Exercise 1.2
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... &=|{{z}_{1}}|\cdot \dfrac{1}{|{{z}_{2}}|}, \text{ by using (1)}\\ &=\dfrac{|{{z}_{2}}|}{|{{z}_{2}}|}=R
- Question 8, Exercise 1.2
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... e{z}=-bi \\ \implies &\overline{z}=-z \quad \text{by using (2).} \end{align} This was required. ====
- Question 1, Exercise 1.1
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)
- Question 2 & 3, Exercise 1.1
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)
- Question 4, Exercise 1.1
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)
- Question 5, Exercise 1.1
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)
- Question 6, Exercise 1.1
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)
- Question 7, Exercise 1.1
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)
- Question 8, Exercise 1.1
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)
- Question 9 & 10, Exercise 1.1
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)