tanθ = $\frac{{\rm{y}}}{{\rm{x}}}$ = $\frac{2}{2}$ = 1, then θ= 45°. So, we can say now, i4n = 1 where n is any positive interger. 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 = 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 $. Similarly, the remainder when f(z) is divided by (z + i) = f(- i) ….. (1), and f( -i) = 1 + i. = cos 60° + i.sin60° = $\frac{1}{2}$ + i.$\frac{{\sqrt 3 }}{2}$ = $\frac{1}{2}$(1 + i$\sqrt 3 $). Complex Numbers extends the concept of one dimensional real numbers to the two dimensional complex numbers in which two dimensions comes from real part and the imaginary part. Either of the part can be zero. = cos13° + i.sin135° = $ - \frac{1}{2}$ + i.$\frac{1}{2}$. Tanθ = $\frac{{ - 1}}{0}$ then θ = 270°. These NCERT Solutions of Maths help the students in solving the problems quickly, accurately and efficiently. Here, z = - 2, y = - 2, r = $\sqrt {4 + 12} $ = 4. the imaginary numbers. Click here for the Detailed Syllabus of IIT JEE Mathematics. = cos(80 – 20) + i.sin(80 – 20) = cos 60° + i.sin 60° = $\frac{1}{2}$ + i.$\frac{{\sqrt 3 }}{2}$. Any equation involving complex numbers in it are called as the complex equation. √a . Question 1. You can assign a value to a complex number in one of the following ways: 1. The sum of four consecutive powers of I is zero.In + in+1 + in+2 + in+3 = 0, n ∈ z 1. A complex number is of the form i 2 =-1. 1/i = – i 2. Tanθ = $\frac{0}{{ - 1}}$ then θ = 180°. A complex number is a number that comprises a real number part and an imaginary number part. Students can also make the best out of its features such as Job Alerts and Latest Updates. 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 . When k = 1, Z1 = cos $\left( {\frac{{180 + 360}}{4}} \right)$ + i.sin $\left( {\frac{{180 + 360}}{4}} \right)$. So, required roots are ± (- 1 + i$\sqrt 3 $). grade, Please choose the valid Main application of complex numbers is in the field of electronics. = $\frac{{\left( {{\rm{cos}}3\theta + {\rm{i}}. a positive and b negative. ‘a’ is called the real part, and ‘b’ is called the imaginary part of the complex number. = cos 225° + i.sin225° = $ - \frac{1}{2}$ + i.$\frac{1}{2}$. FAQ's | This is termed the algebra of complex numbers. NCERT Solutions For Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations are prepared by the expert teachers at BYJU’S. , 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]$. Here, x = -1, y = 0, r = $\sqrt {1 + 0} $ = 1. and if a = 0, z = ib which is called as the Purely Imaginary Number. Tanθ = $\frac{{\rm{y}}}{{\rm{x}}}$ = $\frac{{\frac{{\sqrt 3 }}{2}}}{{\frac{1}{2}}}$ = $\sqrt 3 $ then θ = 60°. The first value represents the real part of the complex number, and the second value represents its imaginary part. 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. 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) Sequence and Series and Mathematical Induction. Q5. Here, x = 0, y = 2, r = $\sqrt {0 + 4} $ = 2. Two mutually perpendicular axes are used to locate any complex point on the plane. Complex Number itself has many ways in which it can be expressed. 1. Tanθ= $\frac{{\rm{y}}}{{\rm{x}}}$ = $\frac{{\sqrt 3 }}{{ - 1}}$ = $ - \sqrt 3 $ then θ = 120°. If ω1 = ω2 are the complex slopes of two lines, then. All the examples listed here are in Cartesian form. {\rm{sin}}\theta } \right)}}{{{{\left( {{\rm{cos}}\theta + {\rm{i}}. The imaginary part, therefore, is a real number! Or, $\sqrt {{{\rm{z}}_{\rm{k}}}} $ = $\sqrt 2 $$\left[ {\cos \frac{{120 + {\rm{k}}.360}}{2} + {\rm{i}}.\sin \frac{{120 + {\rm{k}}.360}}{2}} \right]$, When k = 0, $\sqrt {{{\rm{z}}_0}} $ = $\sqrt 2 $$\left[ {\cos \left( {\frac{{120 + 0}}{2}} \right) + {\rm{i}}.\sin \left( {\frac{{120 + 0}}{2}} \right)} \right]$. Find important formulae and previous year questions related to Complex Numbers for JEE Main and JEE Advanced 2019. Complex Numbers Class 11 solutions NCERT PDF are beneficial in several ways. Or, $\frac{{\rm{i}}}{{1 + {\rm{i}}}}$ = $\frac{{\rm{i}}}{{1 + {\rm{i}}}}$ * $\frac{{1 - {\rm{i}}}}{{1 - {\rm{i}}}}$ = $\frac{{{\rm{i}} - {{\rm{i}}^2}}}{{{1^2} - {{\rm{i}}^2}}}$ = $\frac{{{\rm{i}} - \left( { - 1} \right)}}{{1\left( { - 1} \right)}}$ = $\frac{{{\rm{i}} + 1}}{2}$ = $\frac{1}{2} + \frac{{\rm{i}}}{2}$. Or, 2 $\left( { - \frac{{\sqrt 3 }}{2} + \frac{{{\rm{i}}.1}}{2}} \right)$ = $ - \sqrt 3 $ + i. Careers | Example: Also, note that i + i2 + i3 + i4 = 0 or in + i2n + i3n + i4n= 0. = 210 [-cos0 + i.sin0] = 210 [-1 + i.0] = - 210. Complex numbers are built on the concept of being able to define the square root of negative one. 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). The set of all the complex numbers are generally represented by ‘C’. Complex Numbers are the numbers which along with the real part also has the imaginary part included with it. Students can also make the best out of its features such as Job Alerts and Latest Updates. 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. By a… Out of which, algebraic or rectangular form is one of the form. Practice JEE Main Important Topics Questions solved by our expert teachers helps to score good marks in IIT JEE Exams. Complex Number can be considered as the super-set of all the other different types of number. Let z4 = $ - \frac{1}{2}$ + $\frac{{{\rm{i}}\sqrt 3 }}{2}$. Ltd. Trigonometric Equations and General Values. Sitemap | So, required roots are ± $\frac{1}{{\sqrt 2 }}$(1 + i), ± $\frac{1}{{\sqrt 2 }}$(1 – i). Complex numbers are often denoted by z. Complex Numbers have wide verity of applications in a variety of scientific and related areas such as electromagnetism, fluid dynamics, quantum mechanics, vibration analysis, cartography and control theory. ©Copyright 2014 - 2021 Khulla Kitab Edutech Pvt. Since in third quadrant both a and b are negative and thus a = -2 and b = -3 in our example. Here, x = 0, y = 8, r = $\sqrt {0 + 64} $ = 8. Home ; Grade 11 ; Mathematics ; Sukunda Pustak Bhawan ; Complex Number. Complex numbers are built on the concept of being able to define the square root of negative one. using askIItians. = 2{cos 150° + i.sin150°} = 2 $\left( { - \frac{{\sqrt 3 }}{2} + {\rm{i}}.\frac{1}{2}} \right)$ = - $\sqrt 3 $ + i. Point z is 7 units in the left and 6 units upwards from the origin. = $\frac{1}{{\sqrt 2 }} - {\rm{i}}.\frac{1}{{\sqrt 2 }}$ = $ - \left( { - \frac{1}{{\sqrt 2 }} + {\rm{i}}.\frac{1}{{\sqrt 2 }}} \right)$. Then find the equation whose roots are a19 and b7. Let g(z) be the quotient and az + b the remainder when g(z) is divided by z2 + 1. Moreover, i is just not to distinguish but also has got some value. This point will be lying 5 units in the right and 6 units downwards. = $\frac{1}{{\sqrt 2 }}$ + i.$\frac{1}{{\sqrt 2 }}$. With the help of the NCERT books, students can score well in the JEE Main entrance exam. 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 $\frac{{120 + 0}}{4}$]. If ‘a’ is the real part and ‘b’ represents imaginary part, then complex number is represented as z = a + ib where i, stands for iota which itself is a square root of negative unity. Hence, the equation becomes x2 – (ω + ω2)x + ω ω2 = 0. So, required roots are ± $\left( {\frac{{\sqrt 3 }}{2} + \frac{1}{2}{\rm{i}}} \right)$, ± $\left( {\frac{1}{2} - \frac{{\sqrt 3 }}{2}{\rm{i}}} \right)$. 6. Find the square roots of … The notion of complex numbers increased the solutions to a lot of problems. ‘a’ is called the real part, and ‘b’ is called the imaginary part of the complex number. 2. All educational material on the website has been prepared by the best teachers having more than 20 years of teaching experience in various schools. Look into the Previous Year Papers with Solutions to get a hint of the kinds of questions asked in the exam. For example, 3+2i, -2+i√3 are complex numbers. When, k = 3, Z3 = cos $\left( {\frac{{0 + 1080}}{6}} \right)$ + i.sin $\left( {\frac{{0 + 1080}}{6}} \right)$, When, k = 4, Z4 = cos $\left( {\frac{{0 + 1440}}{6}} \right)$ + i.sin $\left( {\frac{{0 + 1440}}{6}} \right)$, = cos240° + i.sin240° = $ - \frac{1}{2} + {\rm{i}}.\frac{{\sqrt 3 }}{2}$, When, k = 5, Z5 = cos $\left( {\frac{{0 + 1800}}{6}} \right)$ + i.sin $\left( {\frac{{0 + 1800}}{6}} \right)$. Or, $\sqrt {{{\rm{z}}_{\rm{k}}}} $ = $\sqrt 4 $$\left[ {\cos \frac{{120 + {\rm{k}}.360}}{2} + {\rm{i}}.\sin \frac{{120 + {\rm{k}}.360}}{2}} \right]$, When k = 0, $\sqrt {{{\rm{z}}_0}} $ = 2 $\left[ {\cos \left( {\frac{{240 + 0}}{2}} \right) + {\rm{i}}.\sin \left( {\frac{{240 + 0}}{2}} \right)} \right]$. 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 (1 + i)2 = 2i and (1 – i)2 = 2i 3. 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}}}}} $ = $\frac{{1 - {\rm{i}}}}{{\sqrt {1 - {{\rm{i}}^2}} }}$ = $\frac{{1 - {\rm{i}}}}{{\sqrt {1 + 1} }}$ = $\frac{1}{{\sqrt 2 }} - \frac{{\rm{i}}}{{\sqrt 2 }}$. Remainder when f(z) is divided by (z – i) = f(i). Class 8; Class 9; Class 10; Grade 11; Grade 12; Grade 11 Mathematics Solution. 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