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). 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. = $\sqrt 2 ${cos 45° + i.sin45°} = $\sqrt 2 $.$\left( {\frac{1}{{\sqrt 2 }} + {\rm{i}}.\frac{1}{{\sqrt 2 }}} \right)$ = 1 + i. {\rm{sin}}3\theta } \right)\left( {{\rm{cos}}\theta  - {\rm{i}}. When k = 2, Z2 = cos $\left( {\frac{{90 + 720}}{3}} \right)$ + i.sin $\left( {\frac{{90 + 720}}{3}} \right)$. Since both a and b are positive, which means number will be lying in the first quadrant. basically the combination of a real number and an imaginary number tanθ = $\frac{{\rm{y}}}{{\rm{x}}}$ = $ - \frac{1}{1}$ = -1  then θ= 315°. Let g(z) be the quotient and az + b the remainder when g(z) is divided by z2 + 1. Hence, the required remainder  = az + b = ½ iz + ½ + i. Sequence and Series and Mathematical Induction. Tutor log in | A complex number is of the form i 2 =-1. So, required roots are ± $\left( {\frac{{\sqrt 3 }}{2} + \frac{1}{2}{\rm{i}}} \right)$, ± $\left( {\frac{1}{2} - \frac{{\sqrt 3 }}{2}{\rm{i}}} \right)$. Similarly, for z = 3+j5, Re(z) = 3 and Im(z) = (5). = 2$\sqrt 2 $[cos30 + i.sin30] = 2$\sqrt 2 $$\left[ {\frac{{\sqrt 3 }}{2} + {\rm{i}}.\frac{1}{2}} \right]$ = $\sqrt 6 $ + i.$\sqrt 2 $. Students looking for NCERT Solutions for Class 11 Maths can download the same from this article. Let 2=−බ ∴=√−බ Just like how ℝ denotes the real number system, (the set of all real numbers) we use ℂ to denote the set of complex numbers. Free PDF download of Class 11 Maths revision notes & short key-notes for Chapter-5 Complex Numbers and Quadratic Equations to score high marks in exams, prepared by expert mathematics teachers from latest edition of CBSE books. Since arg(a + ib) = π/4, so tan π/4 = b/a which gives a = b, So, 6x2 + 6y2 – 36x – 24y + 66 = 12x – 12y -12, Again, |z – 3 + i| = 3 gives |x + iy - 3 + i| = 3, This yields x2 + y2 - 6x + 2y +1 = 0 …. Ltd. Trigonometric Equations and General Values. The imaginary part, therefore, is a real number! The real part of the complex number is represented by x, and the imaginary part of the complex number is represented by y. If z = -2 + j4, then Re(z) = -2 and Im(z) = 4. Also browse for more study materials on Mathematics here. name, Please Enter the valid = cos 60° + i.sin60° = $\frac{1}{2}$ + i. = $\sqrt 2 $$\left[ { - \frac{1}{2} - {\rm{i}}.\frac{{\sqrt 3 }}{2}} \right]$ = $ - \left( {\frac{1}{{\sqrt 2 }} + \frac{{{\rm{i}}\sqrt 3 }}{{\sqrt 2 }}} \right)$. When k = 2, Z2 = cos $\left( {\frac{{180 + 720}}{4}} \right)$ + i.sin $\left( {\frac{{180 + 720}}{4}} \right)$. Thus, we can also write z = Re(z) + i Im(z). The set of all the complex numbers are generally represented by ‘C’. Find the square roots of … If ω1 = ω2 are the complex slopes of two lines, then. 2 + i3, -5 + 6i, 23i, (2-3i), (12-i1), 3i are some of the examples of complex numbers. When k = 1, $\sqrt {{{\rm{z}}_1}} $ = 2$\sqrt 2 $$\left[ {\cos \left( {\frac{{60 + 360}}{2}} \right) + {\rm{i}}.\sin \left( {\frac{{60 + 360}}{2}} \right)} \right]$. Terms & Conditions | ..... (2). = (cos 32° + i.sin32°)(cos13° + i.sin13°), = cos (32° + 13°) + i.sin(32° + 13°) = cos 45° + i.sin45°. The first value represents the real part of the complex number, and the second value represents its imaginary part. = 2{cos 120° + i.sin120°} = 2.$\left( { - \frac{1}{2} + {\rm{i}}.\frac{{\sqrt 3 }}{2}} \right)$ = $ - {\rm{\: }}1{\rm{\: }}$+ i$\sqrt 3 $. Chapter List. 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. a) Find b and c b) Write down the second root and check it. 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 . = + ∈ℂ, for some , ∈ℝ Complex Numbers Class 11 solutions NCERT PDF are beneficial in several ways. CBSE Worksheets for Class 11 Maths: One of the best teaching strategies employed in most classrooms today is Worksheets. Solving (6) and (7), we have b = ½ + i and a = i/2. Pay Now | Here, x = 1, y = 1, r = $\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} $ = $\sqrt {1 + 1} $ = $\sqrt 2 $. = $\frac{1}{{\sqrt 2 }}$ + i.$\frac{1}{{\sqrt 2 }}$. 2 + i3, -5 + 6i, 23i, (2-3i), (12-i1), 3i are some of the examples of complex numbers. news feed!”. 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. 1/i = – i 2. All the examples listed here are in Cartesian form. Sakshi EAMCET is provided by Sakshieducation.com. 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. a positive and b negative. Chapters. When k = 1, Z1 = 2 {cos$\left( {\frac{{90 + 360}}{3}} \right)$ + i.sin $\left( {\frac{{90 + 360}}{3}} \right)$}. Back to Solutions. How do we locate any Complex Number on the plane? Benefits of Complex Numbers Class 11 NCERT PDF. By calling the static (Shared in Visual Basic) Complex.FromPolarCoordinatesmethod to create a complex number from its polar coordinates. The well-structured Intermediate portal of sakshieducation.com provides study materials for Intermediate, EAMCET.Engineering and Medicine, JEE (Main), JEE (Advanced) and BITSAT. tanθ = $\frac{{\rm{y}}}{{\rm{x}}}$ = $\frac{1}{1}$ = 1 then θ= 45°. 1. Tanθ = $\frac{{ - 1}}{0}$  then θ = 270°. FAQ's | = 2{cos 150° + i.sin150°} = 2 $\left( { - \frac{{\sqrt 3 }}{2} + {\rm{i}}.\frac{1}{2}} \right)$ = - $\sqrt 3 $ + i. Are all Real Numbers are Complex Numbers? (a) If ω1 = ω2 then the lines are parallel. 6. 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. = (cos 30° + i.sin30°) = $\frac{{\sqrt 3 }}{2} + {\rm{i}}.\frac{1}{2}$. tanθ = $\frac{{\rm{y}}}{{\rm{x}}}$ = $\frac{{\frac{1}{2}}}{{\frac{1}{2}}}$ = 1  then θ= 45°. {\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}$]. The well-structured Intermediate portal of sakshieducation.com provides study materials for Intermediate, EAMCET.Engineering and Medicine, JEE (Main), JEE (Advanced) and BITSAT. ‘z’ will be 6 units in the right and 4 units upwards from the origin. √a . Franchisee | Question 1. A complex number is of the form i 2 =-1. This is termed the algebra of complex numbers. When k = 1, $\sqrt {{{\rm{z}}_1}} $ = $\sqrt 2 $$\left[ {\cos \left( {\frac{{120 + 360}}{2}} \right) + {\rm{i}}.\sin \left( {\frac{{120 + 360}}{2}} \right)} \right]$. Also, note that i + i2 + i3 + i4 = 0 or in + i2n + i3n + i4n= 0. Preparing for entrance exams? the imaginary numbers. number, Please choose the valid If z is purely real negative complex number then. (7). So, required roots are ± $\frac{1}{{\sqrt 2 }}$(1 + i), ± $\frac{1}{{\sqrt 2 }}$(1 – i). Let us take few examples to understand that, how can we locate any point on complex or argand plane? 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 $$\left[ {\cos \left( {\frac{{60 + 0}}{2}} \right) + {\rm{i}}.\sin \left( {\frac{{60 + 0}}{2}} \right)} \right]$. = cos 60° + i.sin60° = $\frac{1}{2}$ + i.$\frac{{\sqrt 3 }}{2}$ = $\frac{1}{2}$(1 + i$\sqrt 3 $). Z = a + ib is the algebraic form in which ‘a’ represents real part and ‘b’ represents imaginary part. Or, zk = r1/4$\left\{ {\cos \frac{{\theta  + {\rm{k}}.360}}{4} + {\rm{i}}.\sin \frac{{\theta  + {\rm{k}}.360}}{4}} \right\}$, = 1$\left\{ {\cos \frac{{120 + {\rm{k}}.360}}{4} + {\rm{i}}.\sin \frac{{120 + {\rm{k}}.360}}{4}} \right\}$, When k = 0, Z0 = [cos $\frac{{120 + 0}}{4}$ + i.sin $\frac{{120 + 0}}{4}$]. SPI 3103.2.1 Describe any number in the complex number system. = 1 (cos315° + i.sin315°). Use Coupon: CART20 and get 20% off on all online Study Material, Complete Your Registration (Step 2 of 2 ), Free webinar on Robotics (Block Chain) Learn to create a Robotic Device Using Arduino. By a… Find every complex root of the following. A complex number is usually denoted by the letter ‘z’. Here, x = $ - \frac{1}{2}$, y = $\frac{{\sqrt 3 }}{2}$, r = $\sqrt {\frac{1}{4} + \frac{3}{4}} $ = 1. tanθ = $\frac{{\frac{{\sqrt 3 }}{2}}}{{ - \frac{1}{2}}}$ = $ - \sqrt 3 $ then θ = 120°. $\left[ {\cos \frac{{180 + 0}}{3} + {\rm{i}}.\sin \frac{{180 + 0}}{3}} \right]$. Refund Policy, Register and Get connected with IITian Mathematics faculty, Please choose a valid = 1 (cos90° + i.sin90°). Also after the chapter, you can get links to Class 11 Maths Notes, NCERT Solutions, Important Question, Practice Papers, etc. There are also different ways of representation for the complex number, which we shall learn in the next section. grade, Please choose the valid Or, $\sqrt {{{\rm{z}}_{\rm{k}}}} $ = $\sqrt {\rm{r}} $$\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}} $ =  $\left[ {\cos \left( {\frac{{270 + 360}}{2}} \right) + {\rm{i}}.\sin \left( {\frac{{270 + 360}}{2}} \right)} \right]$. = 210 [-cos0 + i.sin0] = 210 [-1 + i.0] = - 210. Privacy Policy | i is called as Iota in Complex Numbers. Find all complex numbers z such that z 2 = -1 + 2 sqrt(6) i. When k = 1, Z1 ={cos$\left( {\frac{{120 + 360}}{4}} \right)$ + i.sin $\left( {\frac{{120 + 360}}{4}} \right)$}. 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). Any equation involving complex numbers in it are called as the complex equation. School Tie-up | = 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. {\rm{sin}}2\theta }}{{{\rm{cos}}2\theta  + {\rm{i}}. Dear tanθ = $\frac{{\rm{y}}}{{\rm{x}}}$ = $ - \frac{1}{{\sqrt 3 }}$ then θ= 150°. = cos90° + i.sin90°. 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]$. So, required roots are ± (- 1 + i$\sqrt 3 $). Experience in various schools 9 ; Class 9 ; Class 9 ; Class 10 ; 11... 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