This is beta site. https://beta.mathcity.org/ 2026-06-05T19:18:04+00:00 https://beta.mathcity.org/ https://beta.mathcity.org/_media/logo.png text/html 2025-01-21T16:13:57+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-1-p8?rev=1737476037&do=diff Question 9 & 10, Exercise 1.1 Solutions of Question 9 & 10 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 9 $\dfrac{\left( 3-2i \right)\left( 2+3i \right)}{\left( 1+2i \right)\left( 2-i \right)}$\begin{align}z&=\dfrac{\left( 3-2i \right)\left( 2+3i \right)}{\left( 1+2i \right)\left( 2-i \right)}\\ &=\dfrac{6+6+9i-4i}{2+2+4i-i}\\ &=\dfrac{12+5i}{4+3… text/html 2025-01-21T16:13:57+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-3-p2?rev=1737476037&do=diff Question 2, Exercise 1.3 Solutions of Question 2 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 2(i) $P(z)$$$P\left( z \right)={{z}^{3}}+6z+20$$$$p\left( z \right)={{z}^{3}}+6z+20$$$(z-a)$$P(z)$$P(a)=0$$z=-2$\begin{align} P(-2)&=(-2)^3+6(-2)+20\\ &=-8-12+20\\ &=0\end{align}$z+2$${{z}^{3}}+6z+20$$$\begin{array}{c|cccc} -2 & 1 & 0 & 6 & 20 \\ & \d… text/html 2025-01-21T16:13:57+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-3-p5?rev=1737476037&do=diff Question 6, Exercise 1.3 Solutions of Question 6 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 6(i) ${{z}^{4}}+{{z}^{2}}+1=0$$$z^4+z^2+1=0$$$$z^4+2z^2+1-z^2=0$$$$( z^2+1 )^2-z^2=0$$$$( z^2+1+z)( z^2+1-z )=0$$$$( z^2+z+1 )( z^2-z+1 )=0$$$$(z^2+z+1 )=0$$$$z=\dfrac{-1\pm \sqrt{1-4}}{2}$$$$z=\dfrac{-1\pm \sqrt{3}i}{2}$$$$(z^2-z+1 )=0$$$$z=\dfrac{1\pm … text/html 2025-01-21T16:13:57+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-1-p5?rev=1737476037&do=diff Question 6, Exercise 1.1 Solutions of Question 6 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 6(i) $\dfrac{4+i}{3+5i}$$a+ib$\begin{align}\dfrac{4+i}{3+5i}&=\dfrac{4+i}{3+5i}\times \dfrac{3-5i}{3-5i}\\ &=\dfrac{\left( 12+5 \right)+\left( 3-20 \right)i}{9-25{{i}^{2}}}\\ &=\dfrac{17-17i}{9+25}\\ &=\dfrac{17}{34}-\dfrac{17}{34}i\\ &=\dfrac{1}{2}-\dfr… text/html 2025-01-21T16:13:57+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-1-p7?rev=1737476037&do=diff Question 8, Exercise 1.1 Solutions of Question 8 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 8(i) $\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}$$a+ib.$\begin{align}&\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}\\ &=\dfrac{\left( 3+2i \right)\left( 1-2i \right)+\left( 2+i \right)\left( 4-i \right)}{\left( 2+i \right)\left( 3+2i \right)}\\ &=\dfrac{\left( 3+4+2i-6i … text/html 2025-01-21T16:13:57+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-2-p6?rev=1737476037&do=diff Question 7, Exercise 1.2 Solutions of Question 7 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 7(i) $\dfrac{2+3i}{5-2i}$\begin{align}&\dfrac{2+3i}{5-2i} \\ =&\dfrac{2+3i}{5-2i}\times \dfrac{5+2i}{5+2i} \quad \text{by rationalizing} \\ =&\dfrac{10-6+15i+4i}{25+4}\\ =&\dfrac{4+19i}{29}\\ =&\dfrac{4}{29}+\dfrac{19}{29}i \end{align}$=\dfrac{4}{29}$$=\… text/html 2025-01-21T16:13:57+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-3-p4?rev=1737476037&do=diff Question 5, Exercise 1.3 Solutions of Question 5 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 5(i) ${{z}^{2}}+z+3=0$$${{z}^{2}}+z+3=0.$$$a=1$$b=1$$c=3$\begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ &=\dfrac{-1\pm \sqrt{{{\left( 1 \right)}^{2}}-4\left( 1 \right)\left( 3 \right)}}{2\left( 1 \right)}\\ &=\dfrac{-1\pm \sqrt{1-12}}{2}\\ &=\…