When k = 1, $\sqrt {{{\rm{z}}_1}}$ = 2 $\left[ {\cos \left( {\frac{{240 + 360}}{2}} \right) + {\rm{i}}.\sin \left( {\frac{{240 + 360}}{2}} \right)} \right]$, = $\sqrt 2 $$\left( {\frac{1}{2} - \frac{{{\rm{i}}\sqrt 3 }}{2}} \right) = 1 - {\rm{i}}\sqrt 3 . NCERT Solutions For Class 11 Maths: The NCERT Class 11 Maths book contains 16 chapters each with their exercises that help students practice the concepts. Or, \sqrt {{{\rm{z}}_{\rm{k}}}} = \sqrt {\rm{r}}$$\left[ {\cos \frac{{\theta + {\rm{k}}.360}}{2} + {\rm{i}}.\sin \frac{{\theta + {\rm{k}}.360}}{2}} \right]$, When k = 0, $\sqrt {{{\rm{z}}_0}}$ = $\sqrt 2$$\left[ {\cos \left( {\frac{{90 + 0}}{2}} \right) + {\rm{i}}.\sin \left( {\frac{{90 + 0}}{2}} \right)} \right]$. Or, 3 $\left( {\frac{1}{2} + \frac{{{\rm{i}}\sqrt 3 }}{2}} \right)$ = $\frac{3}{2}$ + $\frac{{{\rm{i}}3\sqrt 3 }}{2}$. Remarks. A complex number is usually denoted by the letter ‘z’. = cos 45° + i.sin45° = $\frac{1}{{\sqrt 2 }}$ + i.$\frac{1}{{\sqrt 2 }}$. Here, x = 0, y = 2, r = $\sqrt {0 + 4}$ = 2. Either of the part can be zero. the imaginary numbers. a positive and b negative. = cos300° + i.sin300° = $\frac{1}{2} - {\rm{i}}.\frac{{\sqrt 3 }}{2}$. Look into the Previous Year Papers with Solutions to get a hint of the kinds of questions asked in the exam. Are all Real Numbers are Complex Numbers? Check the below NCERT MCQ Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations with Answers Pdf free download. = (cos 30° + i.sin30°) = $\frac{{\sqrt 3 }}{2} + {\rm{i}}.\frac{1}{2}$. They will get back to you in case of doubts and clear that off in a very efficient manner. Students looking for NCERT Solutions for Class 11 Maths can download the same from this article. Students can also make the best out of its features such as Job Alerts and Latest Updates. {\rm{sin}}(\theta  + {\rm{k}}.360\} $, Or, zk = r1/3$\left\{ {\cos \frac{{\theta  + {\rm{k}}.360}}{3} + {\rm{i}}.\sin \frac{{\theta  + {\rm{k}}.360}}{3}} \right\}$, = 81/3$\left\{ {\cos \frac{{90 + {\rm{k}}.360}}{3} + {\rm{i}}.\sin \frac{{90 + {\rm{k}}.360}}{3}} \right\}$, When k = 0, Z0 = 2 [cos$\frac{{90 + 0}}{3}$+ i.sin$\frac{{90 + 0}}{3}$]. Home ; Grade 11 ; Mathematics ; Sukunda Pustak Bhawan ; Complex Number. {\rm{sin}}3\theta } \right)\left( {{\rm{cos}}\theta - {\rm{i}}. An imaginary number I (iota) is defined as √-1 since I = x√-1 we have i2 = –1 , 13 = –1, i4 = 1 1. By passing two Doublevalues to its constructor. Terms & Conditions | Q5. Sequence and Series and Mathematical Induction. r =$\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} $=$\sqrt {{2^2} + {2^2}} $=$\sqrt {4 + 4} $= 2$\sqrt 2 $. the imaginary numbers. If a = a + bi is a complex number, then a is called its real part, notation a = Re(a), and b is called its imaginary part, notation b = Im(a). Then find the equation whose roots are a19 and b7. Here x =$\frac{1}{2}$, y =$\frac{1}{2}$. Contact Us | = cos 300° + i.sin300° =$\frac{1}{2}$- i.$\frac{{\sqrt 3 }}{2}$=$\frac{{1 - {\rm{i}}\sqrt 3 }}{2}$. ..... (2). SPI 3103.2.1 Describe any number in the complex number system. √b = √ab is valid only when atleast one of a and b is non negative. ©Copyright 2014 - 2021 Khulla Kitab Edutech Pvt. = cos(18 * 5) + i.sin(18 * 5) = cos90° + i.sin90° = 0 + i.1 = i, = cos(9 * 40) + i.sin(9 * 40) = cos 360° + i.sin360° = 1 + i.0 = 1. Illustration 3: Find all complex numbers z for which arg [(3z-6-3i)/(2z-8-6i)] = π/4 and |z-3+4i| = 3. Now let’s consider a point in the third quadrant as z = -2 – j3. That means complex numbers contains two different information included in it. Free PDF download of Important Questions with solutions for CBSE Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations prepared by expert Maths teachers from latest edition of CBSE(NCERT) books. Careers | Back to Solutions. tanθ =$\frac{{\rm{y}}}{{\rm{x}}}$=$\frac{2}{2}$= 1, then θ= 45°. = cos 60° + i.sin60° =$\frac{1}{2}$+ i. Ltd. Trigonometric Equations and General Values. Solved and explained by expert mathematicians. All the examples listed here are in Cartesian form. (c) If ω1 = ω2 then the lines are not parallel. news feed!”. Find every complex root of the following. All educational material on the website has been prepared by the best teachers having more than 20 years of teaching experience in various schools. If z is purely real negative complex number then. Find every complex root of the following. You can assign a value to a complex number in one of the following ways: 1. NCERT Books Free PDF Download for Class 11th to 12th "NCERT Books Free PDF Download for Class 11th to 12th" plays an important role in the JEE Main Preparation because mostly the questions in the JEE Main exam will be asked directly from the NCERT books. = cos 60° + i.sin60° =$\frac{1}{2}$+ i.$\frac{{\sqrt 3 }}{2}$=$\frac{1}{2}$(1 + i$\sqrt 3 $). All the examples listed here are in Cartesian form.$\left( {{\rm{\bar z}}} \right)$= 2π – Arg(z). tanθ =$\frac{{\rm{y}}}{{\rm{x}}}$=$\frac{1}{{ - 1}}$= - 1 then θ= 135°, Z14 = [$\sqrt 2 $(cos 135° + i.sin135°)]14, =${\left( {\sqrt 2 } \right)^{14}}$[cos(135 * 14) + i.sin (135 * 14)], = 27 [cos(90 * 21 + 0) + i.sin(90 * 21 + 0)], = 27 [sin0 + i.cos0] = 27 [0 + i.1] = 27.i, Let z =$\left( {\frac{1}{2} + {\rm{i}}.\frac{{\sqrt 3 }}{2}} \right)$, Here, x =$\frac{1}{2}$, y =$\frac{{\sqrt 3 }}{2}$, r =$\sqrt {{{\left( {\frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}} $= 1. which means i can be assumed as the solution of this equation. When k = 1, Z1 ={cos$\left( {\frac{{120 + 360}}{4}} \right)$+ i.sin$\left( {\frac{{120 + 360}}{4}} \right)$}. = (cos 32° + i.sin32°)(cos13° + i.sin13°), = cos (32° + 13°) + i.sin(32° + 13°) = cos 45° + i.sin45°. Register online for Maths tuition on Vedantu.com to … = - (- 1 + i$\sqrt 3 $). 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