This is beta site. https://beta.mathcity.org/ 2026-06-07T14:03:28+00:00 https://beta.mathcity.org/ https://beta.mathcity.org/_media/logo.png text/html 2025-01-21T16:13:59+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit01/ex1-3-p4?rev=1737476039&do=diff Question 4, Exercise 1.3 Solutions of Question 4 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. Question 4(i) $(1-i) z+(1+i) \omega=3 ; 2 z-(2+5 i) \omega=2+3 i$\begin{align} &(1-i) z+(1+i) \omega=3 \quad \cdots(1)\\ &2 z-(2+5 i) \omega=2+3i \quad\cdots(2) \end{align}$2$\begin{align} &(2-2i)z+(2+2i) \omega=6 \quad \cdots (3) \end{align}$(1-i)$\b… text/html 2025-01-21T16:13:59+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit01/ex1-2-p9?rev=1737476039&do=diff Question 9, Exercise 1.2 Solutions of Question 9 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. Question 9(i) $(2+4 i)^{-1}$$z=2+4i$\begin{align} Re(2+4i)^{-1} & = Re(z^{-1}) = \dfrac{Re(z)}{|z|^2} \\ & =\dfrac{2}{2^2+4^2} = \dfrac{2}{20}\\ &= \dfrac{1}{10}. \end{align}\begin{align} Im(2+4i)^{-1} & = Im(z^{-1}) = -\dfrac{Im(z)}{|z|^2} \\ & =-\df… text/html 2025-01-21T16:13:59+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit01/ex1-4-p1?rev=1737476039&do=diff Question 1, Exercise 1.4 Solutions of Question 1 of Exercise 1.4 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. Question 1(i) $2+i 2 \sqrt{3}$$z=x+iy=2 + i 2 \sqrt{3}$\begin{align} r & = \sqrt{x^2 + y^2} = \sqrt{2^2 + (2\sqrt{3})^2} \\ & = \sqrt{4 + 12} = \sqrt{16} = 4. \end{align}\begin{align} \alpha & = \tan^{-1}\left|\frac{y}{x}\right| = \tan^{-1}\left|\fra… text/html 2025-01-21T16:13:59+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit01/ex1-1-p7?rev=1737476039&do=diff Question 7, Exercise 1.1 Solutions of Question 7 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. Question 7(i) $11+12 i$$$z=11+12i$$\begin{align}|z|&= \sqrt{(11)^2+(12)^2}\\ &=\sqrt{265}\end{align}$|11+12 i|=\sqrt{265}$$(2+3 i)-(2+6 i)$$z=(2+3i)−(2+6i)$\begin{align}z&=2+3i−2−6i\\ &=-3i \end{align}\begin{align} |z| &= \sqrt{0^2+(-3)^2} \\ &= \sqrt{… text/html 2025-01-21T16:13:59+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit01/ex1-1-p2?rev=1737476039&do=diff Question 2, Exercise 1.1 Solutions of Question 2 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. Question 2(i) $x+iy$$(3+2i)+(2+4i)$\begin{align}&(3+i2)+(2+i4)\\ =&(3+2)+(i2+i4)\\ =&5+i6\end{align}$x+iy$$(4+3i)-(2+5i)$\begin{align}&(4+3i)-(2+5i)\\ =&(4-2)+(3i-5i)\\ =&2-2i\end{align}$x+iy$$(4+7i)+(4-7i)$\begin{align} &(4+7i)+(4-7i)\\ =&(4+4)+(7i-7i… text/html 2025-01-21T16:13:59+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit01/ex1-4-p10?rev=1737476039&do=diff Question 9, Exercise 1.4 Solutions of Question 9 of Exercise 1.4 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. Question 9(i) $x=2+3 i$$x_{\max }=1+4 i$$\mathrm{t}=0$$$x=2+3i$$$$x_{\max}=1+4 i$$$$\implies x=x_{\max} e^{i\theta}$$$$2+3i=(1+4 i) e^{i\theta}$$\begin{align} \implies e^{i\theta}&=\dfrac{2+3i}{1+4i} \\ &=\dfrac{(2+3i)(1-4i)}{(1+4i)(1-4i)} \\ &=\dfrac{…