The well-structured Intermediate portal of sakshieducation.com provides study materials for Intermediate, EAMCET.Engineering and Medicine, JEE (Main), JEE (Advanced) and BITSAT. When, k = 3, Z3 = cos $\left( {\frac{{180 + 1080}}{4}} \right)$ + i.sin $\left( {\frac{{180 + 1080}}{4}} \right)$. Here, x = 0, y = 1, r = $\sqrt {0 + 1} $ = 1. 1. Or, 2 $\left( { - \frac{{\sqrt 3 }}{2} + \frac{{{\rm{i}}.1}}{2}} \right)$ = $ - \sqrt 3 $ + i. = cos 300° + i.sin300° = $\frac{1}{2}$ - i.$\frac{{\sqrt 3 }}{2}$ = $\frac{{1 - {\rm{i}}\sqrt 3 }}{2}$. and if a = 0, z = ib which is called as the Purely Imaginary Number. r = $\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} $ = $\sqrt {{2^2} + {2^2}} $ = $\sqrt {4 + 4} $ = 2$\sqrt 2 $. Updated to latest CBSE syllabus. The real part of the complex number is represented by x, and the imaginary part of the complex number is represented by y. Students can also make the best out of its features such as Job Alerts and Latest Updates. When k = 1, Z1 = 2 {cos$\left( {\frac{{90 + 360}}{3}} \right)$ + i.sin $\left( {\frac{{90 + 360}}{3}} \right)$}. Thus we can say that all real numbers are also complex number with imaginary part zero. = 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. Pay Now | tanθ = $\frac{{\rm{y}}}{{\rm{x}}}$ = $ - \frac{1}{1}$ = -1  then θ= 315°. You can assign a value to a complex number in one of the following ways: 1. Dear 2 + i3, -5 + 6i, 23i, (2-3i), (12-i1), 3i are some of the examples of complex numbers. Refund Policy, Register and Get connected with IITian Mathematics faculty, Please choose a valid Complex numbers often are denoted by the letter z or by Greek letters like a (alpha). Contact Us | 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. 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°. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. Tutor log in | , = cos 120° + i.sin120° = $ - \frac{1}{2}$ + i.$\frac{{\sqrt 3 }}{2}$. The first value represents the real part of the complex number, and the second value represents its imaginary part. = $\sqrt 2 ${cos 45° + i.sin45°} = $\sqrt 2 $.$\left( {\frac{1}{{\sqrt 2 }} + {\rm{i}}.\frac{1}{{\sqrt 2 }}} \right)$ = 1 + i. For example, 3+2i, -2+i√3 are complex numbers. $\left[ {\cos \frac{{180 + 0}}{3} + {\rm{i}}.\sin \frac{{180 + 0}}{3}} \right]$. ..... (2). CBSE Class 11 Maths Worksheet for students has been used by teachers & students to develop logical, lingual, analytical, and problem-solving capabilities. Check the below NCERT MCQ Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations with Answers Pdf free download. Two mutually perpendicular axes are used to locate any complex point on the plane. A complex number is usually denoted by the letter ‘z’. School Tie-up | Complex numbers are often denoted by z. When k = 2, Z2 = cos $\left( {\frac{{180 + 720}}{4}} \right)$ + i.sin $\left( {\frac{{180 + 720}}{4}} \right)$. When k = 0, Z0 = 11/6 [cos 0 + i.sin0] = 1. Hence, the required remainder  = az + b = ½ iz + ½ + i. = 1 (cos315° + i.sin315°). “Relax, we won’t flood your facebook = + ∈ℂ, for some , ∈ℝ = $\frac{1}{{\sqrt 2 }}$ (cos45° + i.sin45°). Look into the Previous Year Papers with Solutions to get a hint of the kinds of questions asked in the exam. Click here for the Detailed Syllabus of IIT JEE Mathematics. Tanθ=${\rm{\: }}\frac{{\rm{y}}}{{\rm{x}}}$ = $\frac{{4\sqrt 3 }}{4}$ = $\sqrt 3 $ then θ = 60°. You can see the same point in the figure below. Since both a and b are positive, which means number will be lying in the first quadrant. All educational material on the website has been prepared by the best teachers having more than 20 years of teaching experience in various schools. (1 + i)2 = 2i and (1 – i)2 = 2i 3. {\rm{sin}}3\theta } \right)\left( {{\rm{cos}}\theta  - {\rm{i}}. (d) If ω1 = ω2 then the lines are perpendicular. The complex number 2 + 4i is one of the root to the quadratic equation x 2 + bx + c = 0, where b and c are real numbers. The well-structured Intermediate portal of sakshieducation.com provides study materials for Intermediate, EAMCET.Engineering and Medicine, JEE (Main), JEE (Advanced) and BITSAT. 2 + i3, -5 + 6i, 23i, (2-3i), (12-i1), 3i are some of the examples of complex numbers. By a… If z is purely real negative complex number then. 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. The notion of complex numbers increased the solutions to a lot of problems. Point z is 7 units in the left and 6 units upwards from the origin. 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. When k = 2, Z2 = cos $\left( {\frac{{0 + 720}}{4}} \right)$ + i.sin $\left( {\frac{{0 + 720}}{4}} \right)$, When, k = 3, Z3 = cos $\left( {\frac{{0 + 1080}}{4}} \right)$ + i.sin $\left( {\frac{{0 + 1080}}{4}} \right)$, So, ${\rm{z}}_{\rm{k}}^6$ = r$\{ \cos \left( {\theta  + {\rm{k}}.360} \right) + {\rm{i}}. 2. Hence, Arg. 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 $$\left( {\frac{1}{2} - \frac{{{\rm{i}}\sqrt 3 }}{2}} \right)$ = $1 - {\rm{i}}\sqrt 3 $. Consider a complex number z = 6 +j4 (‘i’ and ‘j’, both can be used for representing imaginary part), if we compare this number with z = a + jb form. $\left[ {\cos \frac{{\theta  + {\rm{k}}.360}}{3} + {\rm{i}}.\sin \frac{{\theta  + {\rm{k}}.360}}{3}} \right]$, Or, ${\rm{z}}_0^{\frac{1}{3}}$ = 1. All the examples listed here are in Cartesian form. Let 2=−බ ∴=√−බ Just like how ℝ denotes the real number system, (the set of all real numbers) we use ℂ to denote the set of complex numbers. √a . (7). It will help you to save your precious time just before the examination. It helps us to clearly distinguish the real and imaginary part of any complex number. 6. addition, multiplication, division etc., need to be defined. Find all complex numbers z such that z 2 = -1 + 2 sqrt(6) i. If a = a + bi is a complex number, then a is called its real part, notation a = Re(a), and b is called its imaginary part, notation b = Im(a). Here, x = $ - \frac{1}{2}$, y = $\frac{{\sqrt 3 }}{2}$, r = $\sqrt {\frac{1}{4} + \frac{3}{4}} $ = 1. tanθ = $\frac{{\frac{{\sqrt 3 }}{2}}}{{ - \frac{1}{2}}}$ = $ - \sqrt 3 $ then θ = 120°. SPI 3103.2.1 Describe any number in the complex number system. These values represent the position of the complex number in the two-dimensional Cartesian coordinate system. = $\sqrt 2 $[cos60° + i.sin60°] = $\sqrt 2 $$\left[ {\frac{1}{2} + {\rm{i}}.\frac{{\sqrt 3 }}{2}} \right]$ = $\frac{1}{{\sqrt 2 }} + \frac{{{\rm{i}}\sqrt 3 }}{{\sqrt 2 }}$. Class 8; Class 9; Class 10; Grade 11; Grade 12; Grade 11 Mathematics Solution. Main application of complex numbers is in the field of electronics. {\rm{sin}}\theta } \right)}^2}}}$, = $\frac{{\left( {{\rm{cos}}3\theta  + {\rm{i}}. Here, x = - 1, y = $\sqrt 3 $, r = $\sqrt {1 + 3} $ = 2. 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. Or, $\sqrt {\frac{{1 - {\rm{i}}}}{{1 + {\rm{i}}}}} $ = $\sqrt {\frac{{1 - {\rm{i}}}}{{1 + {\rm{i}}}}{\rm{*}}\frac{{1 - {\rm{i}}}}{{1 - {\rm{i}}}}} $ = $\frac{{1 - {\rm{i}}}}{{\sqrt {1 - {{\rm{i}}^2}} }}$ = $\frac{{1 - {\rm{i}}}}{{\sqrt {1 + 1} }}$ = $\frac{1}{{\sqrt 2 }} - \frac{{\rm{i}}}{{\sqrt 2 }}$. Here x =$\frac{1}{{\sqrt 2 }}$, y = $ - \frac{1}{{\sqrt 2 }}$. Z = a + ib is the algebraic form in which ‘a’ represents real part and ‘b’ represents imaginary part. With the help of the NCERT books, students can score well in the JEE Main entrance exam. Or, $\frac{1}{{{{\left( {\rm{z}} \right)}^{\rm{n}}}}}$ = z-n = (cosθ + i.sinθ)-n = cos(-n)θ + i.sin(-n)θ, Now, zn – $\frac{1}{{{{\rm{z}}^{\rm{n}}}}}$ = cosnθ + i.sinnθ – cosnθ + i.sinnθ. When k = 1, $\sqrt {{{\rm{z}}_1}} $ = $\sqrt 2 $$\left[ {\cos \left( {\frac{{90 + 360}}{2}} \right) + {\rm{i}}.\sin \left( {\frac{{90 + 360}}{2}} \right)} \right]$. 3. Here you can read Chapter 5 of Class 11 Maths NCERT Book. Thus, we can also write z = Re(z) + i Im(z). 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 | r = $\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} $ = $\sqrt {1 + 1} $ = $\sqrt 2 $. a positive and b negative. Use Coupon: CART20 and get 20% off on all online Study Material, Complete Your Registration (Step 2 of 2 ), Free webinar on Robotics (Block Chain) Learn to create a Robotic Device Using Arduino. (b) If ω1 + ω2 = 0 then the lines are parallel. Complex Numbers Class 11 solutions NCERT PDF are beneficial in several ways. Register yourself for the free demo class from = (sin 40° + i.cos40°)(cos 40° + i.sin40°), = {sin(90° - 50°) + i.cos (90° - 50°)}(cos40° + i.sin40°), = cos(50° + 40°) + i.sin(50° + 40°) = cos 90° + i.sin 90°, = $\frac{{{\rm{cos}}80\infty  + {\rm{i}}. Since in third quadrant both a and b are negative and thus a = -2 and b = -3 in our example. Natural Numbers: Whole Numbers: Integers: Rational Numbers: Irrational Numbers: Types of Rational Numbers: Terminating Decimal Fractions; Recurring and Non-terminating Decimal Fractions: Concept of Radicals and Radicands: Base and Exponent: Definition of a Complex Number: Conjugate of a Complex Number: Section 2: (Exercise No : 2.1) {\rm{sin}}\theta } \right)}}{{{{\left( {{\rm{cos}}\theta  + {\rm{i}}. = cos 60° + i.sin60° = $\frac{1}{2}$ + i.$\frac{{\sqrt 3 }}{2}$ = $\frac{1}{2}$(1 + i$\sqrt 3 $). Here x =$\frac{1}{2}$, y = $\frac{1}{2}$. = 210 [-cos0 + i.sin0] = 210 [-1 + i.0] = - 210. Any equation involving complex numbers in it are called as the complex equation. Concepts of complex numbers, addition, subtraction, multiplication, division of complex numbers. You can get the knowledge of Recommended Books of Mathematics here. Tanθ = $\frac{0}{{ - 1}}$  then θ = 180°. So, required roots are ± $\left( {\frac{1}{{\sqrt 2 }} + \frac{{{\rm{i}}\sqrt 3 }}{{\sqrt 2 }}} \right)$ = ± $\frac{1}{{\sqrt 2 }}$ (1 + i$\sqrt 3 $). ©Copyright 2014 - 2021 Khulla Kitab Edutech Pvt. Here, x = 4, y = 4$\sqrt 3 $, r = $\sqrt {{4^2} + {{\left( {4\sqrt 3 } \right)}^2}} $ = $\sqrt {16 + 48} $ = 8. number, Please choose the valid Free PDF Download of JEE Main Complex Numbers and Quadratic Equations Important Questions of key topics. Sequence and Series and Mathematical Induction. So, Zk = r  [cos (θ + k.360) + i.sin(θ + k.360)], Or, ${\rm{z}}_{\rm{k}}^{\frac{1}{2}}$ = [8{cos (60 + k.360) + i.sin (60 + k.360)}]1/2, = 81/2$\left[ {\cos \left( {\frac{{60 + {\rm{k}}.360}}{2}} \right) + {\rm{i}}.\sin \left( {\frac{{60 + {\rm{k}}.360}}{2}} \right)} \right]$, When k = 0, $\sqrt {{{\rm{z}}_0}} $ = 2$\sqrt 2 $$\left[ {\cos \left( {\frac{{60 + 0}}{2}} \right) + {\rm{i}}.\sin \left( {\frac{{60 + 0}}{2}} \right)} \right]$. which means i can be assumed as the solution of this equation. r = $\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} $ = $\sqrt {\frac{1}{4} + \frac{1}{4}} $ = $\frac{1}{{\sqrt 2 }}$. = (cos 30° + i.sin30°) = $\frac{{\sqrt 3 }}{2} + {\rm{i}}.\frac{1}{2}$. {\rm{sin}}2\theta }}{{{\rm{cos}}2\theta  + {\rm{i}}. Now let’s consider a point in the third quadrant as z = -2 – j3. Then find the equation whose roots are a19 and b7. By passing two Doublevalues to its constructor. √b = √ab is valid only when atleast one of a and b is non negative. Moreover, i is just not to distinguish but also has got some value. = 1 (cos90° + i.sin90°). 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. Prepared by the best teachers with decades of experience, these are the latest Class 11 Maths solutions that you will find. Detailed equations and theorems. Complex numbers are built on the concept of being able to define the square root of negative one. A complex number is usually denoted by the letter ‘z’. Horizontal axis represents real part while the vertical axis represents imaginary part. Having introduced a complex number, the ways in which they can be combined, i.e. CBSE Class 11 Mathematics Worksheet - Complex Numbers and Quadratic Equation (1) CBSE,CCE and NCERT students can refer to the attached file. If ω1 = ω2 are the complex slopes of two lines, then. Let us have a look at the types of questions asked in the exam from this topic: Illustration 1: Let a and b be roots of the equation x2 + x + 1 = 0. 2cos45° - i.2sin45° = 2.$\frac{1}{{\sqrt 2 }}$ – i.2.$\frac{1}{{\sqrt 2 }}$ = $\sqrt 2 $ – i$\sqrt 2 $. This means sum of consecutive four powers of iota leads the result to zero. Yes of course, but to understand this question, let’s go into more deep of complex numbers, Consider the equation x2+1 = 0, If we try to get its solution, we would stuck at x = √(-1) so in Complex Number we assume that ​√(-1) =i or i2 =-1. Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.5 Additional Problems. = $\sqrt 2 $$\left( { - \frac{1}{{\sqrt 2 }} - \frac{{\rm{i}}}{{\sqrt 2 }}} \right)$ =  - 1 – i = - (1 + i). Or, $\sqrt {{{\rm{z}}_{\rm{k}}}} $ = $\sqrt {\rm{r}} $$\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}} $ =  $\left[ {\cos \left( {\frac{{270 + 360}}{2}} \right) + {\rm{i}}.\sin \left( {\frac{{270 + 360}}{2}} \right)} \right]$. Or, 3 $\left( { - \frac{1}{2} + \frac{{{\rm{i}}\sqrt 3 }}{2}} \right)$ = $ - \frac{3}{2}$ + $\frac{{{\rm{i}}3\sqrt 3 }}{2}$. Illustration 3: Find all complex numbers z for which arg [(3z-6-3i)/(2z-8-6i)] = π/4 and |z-3+4i| = 3. Again, ${\rm{\bar z}}$ = r(cosθ – i.sinθ) = r[cos (2π – θ) + i.sin(2π – θ)], So, Arg $\left( {{\rm{\bar z}}} \right)$ = 2π – θ = 2π – Arg (z). Find important formulae and previous year questions related to Complex Numbers for JEE Main and JEE Advanced 2019. = cos 45° + i.sin45° = $\frac{1}{{\sqrt 2 }}$ + i.$\frac{1}{{\sqrt 2 }}$. z = -7 + j6, Here since a= -7 and b = 6 and thus will be lying in the second quadrant. Ltd. Trigonometric Equations and General Values. Find the remainder upon the division of f(z) by z2 + 1. ‘i’ (or ‘j’ in some books) in math is used to denote the imaginary part of any complex number. = (cos315° + i.sin315°). news feed!”. Back to Solutions. {\rm{sin}}\theta } \right)}^2}}}$, = $\frac{{{\rm{cos}}2\theta  + {\rm{i}}. FAQ's | Click Here to Download Mathematics Formula Sheet pdf 4. (b) If z = a + ib is the complex number, then a and b are called real and imaginary parts, respectively, of the complex number and written as R e (z) = a, Im (z) = b. A complex number is a number that comprises a real number part and an imaginary number part. tanθ = $\frac{{\rm{y}}}{{\rm{x}}}$ = $\frac{1}{1}$ = 1 then θ= 45°. Find the square roots of … Sakshi EAMCET is provided by Sakshieducation.com. Email, Please Enter the valid mobile using askIItians. tanθ = $\frac{{\rm{y}}}{{\rm{x}}}$ = $\frac{{\frac{1}{2}}}{{\frac{1}{2}}}$ = 1  then θ= 45°. = cos 225° + i.sin225° = $ - \frac{1}{2}$ + i.$\frac{1}{2}$. tanθ = $\frac{{\rm{y}}}{{\rm{x}}}$ = $ - \frac{1}{{\sqrt 3 }}$ then θ= 150°. (a) If ω1 = ω2 then the lines are parallel. Privacy Policy | {\rm{sin}}3\theta } \right)\left\{ {\cos \left( { - \theta } \right) + {\rm{i}}.\sin \left( { - \theta } \right)} \right\}}}{{{{\left( {{\rm{cos}}\theta  + {\rm{isin}}\theta } \right)}^2}}}$, = $\frac{{\cos \left( {3\theta  - \theta } \right) + {\rm{i}}.\sin \left( {3\theta  - \theta } \right)}}{{{{\left( {{\rm{cos}}\theta  + {\rm{i}}. Since arg(a + ib) = π/4, so tan π/4 = b/a which gives a = b, So, 6x2 + 6y2 – 36x – 24y + 66 = 12x – 12y -12, Again, |z – 3 + i| = 3 gives |x + iy - 3 + i| = 3, This yields x2 + y2 - 6x + 2y +1 = 0 …. = 2(cos 30° + i.sin30°) = $2\left( {\frac{{\sqrt 3 }}{2} + {\rm{i}}.\frac{1}{2}} \right)$ = $\sqrt 3 $ + i. {\rm{sin}}80\infty }}{{{\rm{cos}}20\infty  + {\rm{i}}. Complex numbers are defined as numbers of the form x+iy, where x and y are real numbers and i = √-1. Chapter List. Digital NCERT Books Class 11 Maths pdf are always handy to use when you do not have access to physical copy. basically the combination of a real number and an imaginary number This chapter provides detailed information about the complex numbers and how to represent complex numbers in the Arg and plane. , then Re ( z ) = ( 5 ) and argument of the NCERT Books Class Maths... On complex or argand plane the set of all the examples listed here are in form! 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