= 2{cos 150° + i.sin150°} = 2 $\left( { - \frac{{\sqrt 3 }}{2} + {\rm{i}}.\frac{1}{2}} \right)$ = - $\sqrt 3$ + i. tanθ = $\frac{{\rm{y}}}{{\rm{x}}}$ = $- \frac{1}{1}$ = -1  then θ= 315°. The well-structured Intermediate portal of sakshieducation.com provides study materials for Intermediate, EAMCET.Engineering and Medicine, JEE (Main), JEE (Advanced) and BITSAT. Find the square roots of … 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). If z = -2 + j4, then Re(z) = -2 and Im(z) = 4. Blog | The complex number in the polar form = r(cosθ + i.sinθ) For a complex number z = x+iy, x is called the real part, denoted by Re z and y is called the imaginary part denoted by Im z. The set of all the complex numbers are generally represented by ‘C’. That means complex numbers contains two different information included in it. Click hereto get an answer to your question ️ For all complex numbers z1, z2 satisfying |z1| = 12 and |z2 - 3 - 4 i| = 5 , the minimum value of |z1 - z2| is Here, x = -1, y = 0, r = $\sqrt {1 + 0}$ = 1. Tanθ=${\rm{\: }}\frac{{\rm{y}}}{{\rm{x}}}$ = $\frac{{4\sqrt 3 }}{4}$ = $\sqrt 3$ then θ = 60°. “Relax, we won’t flood your facebook (c) If ω1 = ω2 then the lines are not parallel. Horizontal axis represents real part while the vertical axis represents imaginary part. Here, x = 4, y = 4$\sqrt 3$, r = $\sqrt {{4^2} + {{\left( {4\sqrt 3 } \right)}^2}}$ = $\sqrt {16 + 48}$ = 8. Free PDF download of Class 11 Maths revision notes & short key-notes for Chapter-5 Complex Numbers and Quadratic Equations to score high marks in exams, prepared by expert mathematics teachers from latest edition of CBSE books. Complex Number can be considered as the super-set of all the other different types of number. ‘a’ is called the real part, and ‘b’ is called the imaginary part of the complex number. Dear Here you can read Chapter 5 of Class 11 Maths NCERT Book. (b) If ω1 + ω2 = 0 then the lines are parallel. So, required roots are ± $\left( {\frac{{\sqrt 3 }}{2} + \frac{1}{2}{\rm{i}}} \right)$, ± $\left( {\frac{1}{2} - \frac{{\sqrt 3 }}{2}{\rm{i}}} \right)$. r = $\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}}$ = $\sqrt {{2^2} + {2^2}}$ = $\sqrt {4 + 4}$ = 2$\sqrt 2$. Tutor log in | Here, x = 0, y = 1, r = $\sqrt {0 + 1}$ = 1. Or, 3 $\left( {\frac{1}{2} + \frac{{{\rm{i}}\sqrt 3 }}{2}} \right)$ = $\frac{3}{2}$ + $\frac{{{\rm{i}}3\sqrt 3 }}{2}$. Express your answer in Cartesian form (a+bi): (a) z3 = i z3 = ei(π 2 +n2π) =⇒ z = ei(π 2 +n2π)/3 = ei(π 6 +n2π 3) n = 0 : z = eiπ6 = cos π 6 +isin π 6 = 3 2 + 1 i n = 1 : z = ei56π = cos 5π 6 +isin 5π Get Important questions for class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations here.Students can get different types of questions covered in this chapter. r = $\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}}$ = $\sqrt {\frac{1}{2} + \frac{1}{2}}$ = 1. tanθ = $\frac{{\rm{y}}}{{\rm{x}}}$ = $\frac{{ - \frac{1}{{\sqrt 2 }}}}{{\frac{1}{{\sqrt 2 }}}}$ = -1  then θ= 315°. ir = ir 1. {\rm{sin}}2\theta }}$= cos (2θ – 2θ) + i.sin(2θ – 2θ). Or,$\sqrt {{{\rm{z}}_{\rm{k}}}} $=$\sqrt {\rm{r}} $$\left[ {\cos \frac{{270 + 0}}{2} + {\rm{i}}.\sin \frac{{270 + 0}}{2}} \right], = [cos 135 + i.sin135] =  - \frac{1}{{\sqrt 2 }} + {\rm{i}}.\frac{1}{{\sqrt 2 }}, When k = 1, \sqrt {{{\rm{z}}_1}}  = \left[ {\cos \left( {\frac{{270 + 360}}{2}} \right) + {\rm{i}}.\sin \left( {\frac{{270 + 360}}{2}} \right)} \right]. ©Copyright 2014 - 2021 Khulla Kitab Edutech Pvt. Free PDF Download of JEE Main Complex Numbers and Quadratic Equations Important Questions of key topics. {\rm{sin}}2\theta }}{{{\rm{cos}}2\theta + {\rm{i}}. = \sqrt 2 {cos 45° + i.sin45°} = \sqrt 2 .\left( {\frac{1}{{\sqrt 2 }} + {\rm{i}}.\frac{1}{{\sqrt 2 }}} \right) = 1 + i. For example: x = (2+3i) (3+4i), In this example, x is a multiple of two complex numbers. (c) 0 if r is not a multiple of 3 and 3 if r is a multiple of 3. Pay Now | Complex Numbers (a + bi) Natural (Counting) Numbers Whole Numbers Integers Rational Numbers Real Numbers Irrational #’s Imaginary #’s Complex Numbers are written in the form a + bi, where a is the real part and b is the imaginary part. = 2\sqrt 2 [cos30 + i.sin30] = 2\sqrt 2$$\left[ {\frac{{\sqrt 3 }}{2} + {\rm{i}}.\frac{1}{2}} \right]$=$\sqrt 6 $+ i.$\sqrt 2 $. =$\sqrt 2 $$\left[ { - \frac{1}{2} - {\rm{i}}.\frac{{\sqrt 3 }}{2}} \right] =  - \left( {\frac{1}{{\sqrt 2 }} + \frac{{{\rm{i}}\sqrt 3 }}{{\sqrt 2 }}} \right). = \frac{1}{{\sqrt 2 }} - {\rm{i}}.\frac{1}{{\sqrt 2 }} =  - \left( { - \frac{1}{{\sqrt 2 }} + {\rm{i}}.\frac{1}{{\sqrt 2 }}} \right). Complex numbers are often denoted by z. When k = 1, Z1 = 2 {cos\left( {\frac{{90 + 360}}{3}} \right) + i.sin \left( {\frac{{90 + 360}}{3}} \right)}. Complex number has two parts, real part and the imaginary part. number, Please choose the valid Or, {\rm{z}}_{\rm{k}}^{\frac{1}{3}} = 1. 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. Register yourself for the free demo class from r = \sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}}  = \sqrt {\frac{1}{4} + \frac{1}{4}}  = \frac{1}{{\sqrt 2 }}. Register Now. i is called as Iota in Complex Numbers. \left[ {\cos \frac{{180 + 0}}{3} + {\rm{i}}.\sin \frac{{180 + 0}}{3}} \right]. You will see that, in general, you proceed as in real numbers, but using i 2 =−1 where appropriate. Here, x = 0, y = 8, r = \sqrt {0 + 64}  = 8. It provides EAMCET Mock tests, Online Practice Tests, EAMCET Bit banks, EAMCET Previous Solved Model Papers, and also it gives you the experts guidance. Refund Policy, Register and Get connected with IITian Mathematics faculty, Please choose a valid 2. Complex numbers as ordered pairs of reals, Representation of complex numbers in the form a+ib and their representation in a plane, Argand diagram, algebra of complex numbers, modulus and argument (or amplitude) of a complex number, square root of a complex number, triangle inequality. Illustration 3: Find all complex numbers z for which arg [(3z-6-3i)/(2z-8-6i)] = π/4 and |z-3+4i| = 3. = cos(80 – 20) + i.sin(80 – 20) = cos 60° + i.sin 60° = \frac{1}{2} + i.\frac{{\sqrt 3 }}{2}. \left[ {\cos \frac{{180 + 720}}{3} + {\rm{i}}.\sin \frac{{180 + 720}}{3}} \right]. Tanθ= \frac{{\rm{y}}}{{\rm{x}}} = \frac{{\sqrt 3 }}{{ - 1}} =  - \sqrt 3  then θ = 120°. Here, x = -1, y = 1, r = \sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}}  = \sqrt {{{\left( { - 1} \right)}^2} + {1^2}}  = \sqrt 2 . The imaginary part, therefore, is a real number! and if a = 0, z = ib which is called as the Purely Imaginary Number. Prepared by the best teachers with decades of experience, these are the latest Class 11 Maths solutions that you will find. It is the exclusive and best Telegu education portal established by Sakshi Media Group. A similar problem was … Sitemap | You can see the same point in the figure below. Some of the most commonly used forms are: Cartesian or algebraic or rectangular form. = \frac{1}{{\sqrt 2 }} (cos45° + i.sin45°). Find the remainder upon the division of f(z) by z2 + 1. Two mutually perpendicular axes are used to locate any complex point on the plane. {\rm{sin}}(\theta + {\rm{k}}.360\} , Or, zk = r1/6\left\{ {\cos \frac{{0 + {\rm{k}}.360}}{4} + {\rm{i}}.\sin \frac{{0 + {\rm{k}}.360}}{4}} \right\}. Home ; Grade 11 ; Mathematics ; Sukunda Pustak Bhawan ; Complex Number. On multiplying these two complex number we can get the value of x. z2 + 2z + 3 = 0 is also an example of complex equation whose solution can be any complex number. So, z = r (cosθ + i.sinθ) = \sqrt 2 (cos 45° + i.sin45°), Or, z20 = [\sqrt 2 (cos 45° + i.sin45°)]20, = {\left( {\sqrt 2 } \right)^{20}}[cos(45 * 20) + i.sin (45 * 20)], = 210 [cos(90 * 10 + 0) + i.sin (90 * 10 + 0)]. So, required roots are ± \frac{1}{{\sqrt 2 }}(1 – i). Refer the figure to understand it pictorially. When k = 1, Z1 = cos \left( {\frac{{0 + 360}}{4}} \right) + i.sin \left( {\frac{{0 + 360}}{4}} \right). The notion of complex numbers increased the solutions to a lot of problems. Media Coverage | Tanθ = \frac{{\rm{y}}}{{\rm{x}}} = \frac{{\frac{{\sqrt 3 }}{2}}}{{\frac{1}{2}}} = \sqrt 3  then θ = 60°. grade, Please choose the valid Or, 2 \left( { - \frac{{\sqrt 3 }}{2} + \frac{{{\rm{i}}.1}}{2}} \right) =  - \sqrt 3  + i. With the help of the NCERT books, students can score well in the JEE Main entrance exam. With the help of the NCERT books, students can score well in the JEE Main entrance exam. (d) If ω1 = ω2 then the lines are perpendicular. Complex numbers are often denoted by z. Check the below NCERT MCQ Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations with Answers Pdf free download. 6. {\rm{sin}}(\theta + {\rm{k}}.360\} , Or, zk = r1/4\left\{ {\cos \frac{{0 + {\rm{k}}.360}}{4} + {\rm{i}}.\sin \frac{{0 + {\rm{k}}.360}}{4}} \right\}. = 210 [-cos0 + i.sin0] = 210 [-1 + i.0] = - 210. = \frac{1}{{\sqrt 2 }} + i.\frac{1}{{\sqrt 2 }}. Or, \frac{{\rm{i}}}{{1 + {\rm{i}}}} = \frac{{\rm{i}}}{{1 + {\rm{i}}}} * \frac{{1 - {\rm{i}}}}{{1 - {\rm{i}}}} = \frac{{{\rm{i}} - {{\rm{i}}^2}}}{{{1^2} - {{\rm{i}}^2}}} = \frac{{{\rm{i}} - \left( { - 1} \right)}}{{1\left( { - 1} \right)}} = \frac{{{\rm{i}} + 1}}{2} = \frac{1}{2} + \frac{{\rm{i}}}{2}. It helps us to clearly distinguish the real and imaginary part of any complex number. Out of which, algebraic or rectangular form is one of the form. Terms & Conditions | {\rm{sin}}\theta } \right)}^2}}}, = \frac{{\left( {{\rm{cos}}3\theta + {\rm{i}}. When k = 2, Z2 = cos \left( {\frac{{0 + 720}}{6}} \right) + i.sin \left( {\frac{{0 + 720}}{6}} \right). All the examples listed here are in Cartesian form. 4. All educational material on the website has been prepared by the best teachers having more than 20 years of teaching experience in various schools. Question 1. i2 = z2 = (cos 90° + i.sin90°)2 = cos(90 * 2) + i.sin(90 * 2) = cos 180° + i.sin180° = - 1. Find the modulus and argument of the following complex numbers and convert them in polar form. Either of the part can be zero. An imaginary number I (iota) is defined as √-1 since I = x√-1 we have i2 = –1 , 13 = –1, i4 = 1 1. Also, note that i + i2 + i3 + i4 = 0 or in + i2n + i3n + i4n= 0. Benefits of Complex Numbers Class 11 NCERT PDF. Register online for Maths tuition on Vedantu.com to … Here, z = 0, y = 1, r = \sqrt {{0^2} + {1^2}}  = 1, So, z = 1(cosθ + i.sinθ) = 1. \frac{{\sqrt 3 }}{2}. By a… What is the application of Complex Numbers? Here, z = - 2, y = - 2, r = \sqrt {4 + 12}  = 4. = 2\sqrt 2$$\left[ { - \frac{{\sqrt 3 }}{2} - {\rm{i}}.\frac{1}{3}} \right]$=$ - \left( {\sqrt 6  + {\rm{i}}.\sqrt 2 } \right)$. Then find the equation whose roots are a19 and b7. Updated to latest CBSE syllabus. So, required roots are$\sqrt 3 $+ i,$ - \sqrt 3 $+ i , - 2i. When k = 0, Z0 = 11/6 [cos 0 + i.sin0] = 1. Similarly, the remainder when f(z) is divided by (z + i) = f(- i) ….. (1), and f( -i) = 1 + i. Or, zk = r1/4$\left\{ {\cos \frac{{\theta  + {\rm{k}}.360}}{4} + {\rm{i}}.\sin \frac{{\theta  + {\rm{k}}.360}}{4}} \right\}$, = 1$\left\{ {\cos \frac{{120 + {\rm{k}}.360}}{4} + {\rm{i}}.\sin \frac{{120 + {\rm{k}}.360}}{4}} \right\}$, When k = 0, Z0 = [cos$\frac{{120 + 0}}{4}$+ i.sin$\frac{{120 + 0}}{4}$]. Since in third quadrant both a and b are negative and thus a = -2 and b = -3 in our example. This point will be lying 5 units in the right and 6 units downwards. Let 2=−බ ∴=√−බ Just like how ℝ denotes the real number system, (the set of all real numbers) we use ℂ to denote the set of complex numbers. To find the value of in (n > 4) first, divide n by 4.Let q is the quotient and r is the remainder.n = 4q + r where o < r < 3in = i4q + r = (i4)q , ir = (1)q . Careers | = 12−8−15+102 9−6+6−42 = 12−23+10(−1) 9−4(−1) =2−23 13 = − Graphical Representation A complex number can be represented on an Argand diagram by plotting the real part on the -axis and the imaginary part on the y-axis. When k = 1, Z1 = cos$\left( {\frac{{180 + 360}}{4}} \right)$+ i.sin$\left( {\frac{{180 + 360}}{4}} \right)$. Express your answer in Cartesian form (a+bi): (a) z3 = i z3 = ei(π 2 +n2π) =⇒ z = ei(π 2 +n2π)/3 = ei(π 6 +n2π 3) n = 0 : z = eiπ6 = cos π 6 +isin π 6 = 3 2 + 1 i n = 1 : z = ei56π = cos 5π 6 +isin 5π = 2 {cos 270° + i.sin270°} = 2{0 + i. When k = 1, Z1 ={cos$\left( {\frac{{120 + 360}}{4}} \right)$+ i.sin$\left( {\frac{{120 + 360}}{4}} \right)$}. To read more, Buy study materials of Complex Numbers comprising study notes, revision notes, video lectures, previous year solved questions etc. Main application of complex numbers is in the field of electronics. = + ∈ℂ, for some , ∈ℝ = (cos 32° + i.sin32°)(cos13° + i.sin13°), = cos (32° + 13°) + i.sin(32° + 13°) = cos 45° + i.sin45°. Complex numbers are defined as numbers of the form x+iy, where x and y are real numbers and i = √-1. = cos 60° + i.sin60° =$\frac{1}{2}$+ i.$\frac{{\sqrt 3 }}{2}$=$\frac{1}{2}$(1 + i$\sqrt 3 $). These NCERT Solutions of Maths help the students in solving the problems quickly, accurately and efficiently. Or,${\rm{z}}_1^{\frac{1}{3}}$=$\left[ {\cos \frac{{180 + 360}}{3} + {\rm{i}}.\sin \frac{{180 + 360}}{3}} \right]$, Or,${\rm{z}}_2^{\frac{1}{3}}$= 1. By passing two Doublevalues to its constructor. Argument of a Complex Number Argument of a... Complex Number System Indian mathematician... n th Roots of Unity In general, the term root of... About Us | Illustration 2: Dividing f(z) by z - i, we obtain the remainder i and dividing it by z + i, we get remainder 1 + i. It will help you to save your precious time just before the examination. = cos 300° + i.sin300° =$\frac{1}{2}$- i.$\frac{{\sqrt 3 }}{2}$=$\frac{{1 - {\rm{i}}\sqrt 3 }}{2}$. We know from the above discussion that, Complex Numbers can be represented in four different ways. By calling the static (Shared in Visual Basic) Complex.FromPolarCoordinatesmethod to create a complex number from its polar coordinates. The imaginary part, therefore, is a real number! = cos 60° + i.sin60° =$\frac{1}{2}$+ i. 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. Grade 12; PRACTICE. Solved and explained by expert mathematicians. 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