An Introduction Any number of the form x + iy where x,y are Real and i=√-1, i.e., i2 = -1 is called a complex number. For example, 7 + i10, -5 -4i are examples of complex numbers. The set or complex numbers is denoted by C, i.e., C={x+iy: x,y belongs to R(set of Real number) and i=√-1}. The complex number x+yi is usually denoted by z; x is called real part of z and is written as Re(z) and y is called imaginary part or z and is written as Im(z). So, for every z belongs to C. Z=Re(z) + i Im(z). For example, if z= 3 +5i, then Re(z)=3 and Im (z)=5. Addition of complex numbers Let z1=x1+iy1 and z2=x2+iy2 be any two complex numbers, then the sum or addition of z1 and z2 is defined as (x1+x2) + i(y1+y2), and it is denoted by z1+z2 . The sum or two or more complex numbers is also a complex number. Negative of a complex number Let z=x+iy be any complex number, then the number –x-iy is called negative of z and is denoted by –z. Negative of a complex number is again a complex number. Difference of Complex numbers Let z1=x1+iy1 and z2=x2+iy2 be any two complex numbers, then the difference of z2 from z1 is defined as (x1-x2) + i(y1-y2), and it is denoted by z1-z2 . The difference of two complex numbers is also a complex number. Multiplication of complex numbers. Let z1=x1+iy1 and z2=x2+iy2 be any two complex numbers, then the product or multiplication of z1 and z2 is defined as (x1x2 - y1y2) + i(x1y2 + y1x2) and it is denoted by z1z2 . For example: z1 = 5+3i and z2 = 2+4i then z1z2= (10-12) +i(20+6)=-2+26i. Lets understand the entire process with the help of an example. Find the square roots of 3-4i Solution: let x+iy be a square root of 3-4i Then (x+iy)2 =3-4i x2-y2 +2xyi=3-4i ( (a+b)2 =a2+b2+2ab) => x2–y2=3 ………………..(i) And 2xy=-4, i.e., xy=-2………..(ii) We have (x2+y2)2= (x2-y2)2+4x2y2 32+4(4)=9+16=25 x2+y2=+5 but as x2+y2 =5………………..(iii) On adding (i) and (iii), we get 2x2 =8 => x2 =4 => x=+2 Substituting in (ii), we get x=2, y=-1 and x=-2, y=1. Therefore, the two square roots of 3-4i are 2-i and Let x be a cube root of unity, then x3=1. => x3-1=0=> (x-1)(x2 +x+1)=0 => either x-1=0 or x2 +x+1=0 Either x=1or x=(-1+√-3)/2 Thus, the three cube roots of unity are 1, (-1+i√3)/2 and (-1-i√3)/2. Either of the two non-real cube roots of unity is the square of the other. If one of the non-real cube root of unity is denoted by w(read it as omega), then the other is w2. Further, to avoid any possible confusion, we shall take w= (-1+i√3)/2 Thus, the three cube roots of unity are 1,w and w2. Sum of the three cube roots of unity is zero. Thus 1+w+w2=0 Product of the cube roots of unity is one. i.e. 1.w.w2 =w3=1 Either of the two non-real cube roots of unity is reciprocal of the other. Since w3=1, therefore, w.w2=1=> w and w2 are reciprocals of each other. When we write the complex number in polar form, then z=r(cos φ + i sin φ) Then according to De-Moivre’s Theorem zn = [r(cos φ + i sin φ)]n = rn(cos nφ + i sin nφ) If the Discriminant of any quadratic equation <0 i.e. –ve number. The roots of that quadratic equation are not real, that means the roots of such quadratic equation are complex number. If d<0 where d=b2-4ac Then roots are (–b+√b2-4ac)/2a. Solve the equation 3x2+7=0 Here, the discriminant=02-4x3x7=-84 Therefore x=(-0+√-84)/2x3 =(+2(√21)i)/6 => (+(√21)i)/3

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# Complex Numbers - Govt. PG College Una (HP)

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