This is beta site. https://beta.mathcity.org/ 2026-06-06T12:07:42+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-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}$$a+ib.$\begin{align}&\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}\\ &=\dfrac{\left( 3+2i \right)\left( 1-2i \right)+\left( 2+i \right)\left( 4-i \right)}{\left( 2+i \right)\left( 3+2i \right)}\\ &=\dfrac{\left( 3+4+2i-6i … text/html 2025-01-21T16:13:57+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-2-p2?rev=1737476037&do=diff Question 2, Exercise 1.2 Solutions of Question 2 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 2 $z_1=-1+i$, $z_2=3-2i$${{z}_{3}}=2-2i$${{z}_{1}}=-1+i$${{z}_{2}}=3-2i$${{z}_{3}}=2-2i$$$(z_1+z_2)+z_3=z_1+(z_2+z_3).$$\begin{align} {{z}_{1}}+{{z}_{2}}&=\left( -1+i \right)+\left( 3-2i \right)\\ &=2-i\end{align}\begin{align} \left( {{z}_{1}}+{{z}_{2}… text/html 2025-01-21T16:13:57+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-1-p8?rev=1737476037&do=diff Question 9 & 10, Exercise 1.1 Solutions of Question 9 & 10 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 9 $\dfrac{\left( 3-2i \right)\left( 2+3i \right)}{\left( 1+2i \right)\left( 2-i \right)}$\begin{align}z&=\dfrac{\left( 3-2i \right)\left( 2+3i \right)}{\left( 1+2i \right)\left( 2-i \right)}\\ &=\dfrac{6+6+9i-4i}{2+2+4i-i}\\ &=\dfrac{12+5i}{4+3… text/html 2025-01-21T16:13:57+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-2-p6?rev=1737476037&do=diff Question 7, Exercise 1.2 Solutions of Question 7 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 7(i) $\dfrac{2+3i}{5-2i}$\begin{align}&\dfrac{2+3i}{5-2i} \\ =&\dfrac{2+3i}{5-2i}\times \dfrac{5+2i}{5+2i} \quad \text{by rationalizing} \\ =&\dfrac{10-6+15i+4i}{25+4}\\ =&\dfrac{4+19i}{29}\\ =&\dfrac{4}{29}+\dfrac{19}{29}i \end{align}$=\dfrac{4}{29}$$=\… 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-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$$$\begin{array}{c|cccc} -2 & 1 & 0 & 6 & 20 \\ & \d… text/html 2025-01-21T16:13:57+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit01/ex1-1-p9?rev=1737476037&do=diff Question 11, Exercise 1.1 Solutions of Question 11 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 11(i) ${{z}_{1}}=2-i$${{z}_{2}}=-2+i$${\rm Re}\left( \dfrac{{{z}_{1}}{{z}_{2}}}{\overline{{{z}_{1}}}} \right)$$z_1=2-i$$z_2=-2+i$$\overline{z_1}=2+i$\begin{align} z_1 z_2&=(2-i)(-2+i)\\ &=-4+1+2i+2i\\ &=-3+4i \end{align}\begin{align} \dfrac{z_1 z_2}{\…