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Published byJeffery Randall Modified over 2 years ago

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An Introduction

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Any number of the form x + iy where x,y are Real and i=-1, i.e., i 2 = -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.

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Addition of complex numbers Let z 1 =x 1 +iy 1 and z 2 =x 2 +iy 2 be any two complex numbers, then the sum or addition of z 1 and z 2 is defined as (x 1 +x 2 ) + i(y 1 +y 2 ), and it is denoted by z 1 +z 2. 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.

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Difference of Complex numbers Let z 1 =x 1 +iy 1 and z 2 =x 2 +iy 2 be any two complex numbers, then the difference of z 2 from z 1 is defined as (x 1 -x 2 ) + i(y 1 -y 2 ), and it is denoted by z 1 -z 2. The difference of two complex numbers is also a complex number.

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Multiplication of complex numbers. Let z 1 =x 1 +iy 1 and z 2 =x 2 +iy 2 be any two complex numbers, then the product or multiplication of z 1 and z 2 is defined as (x 1 x 2 - y 1 y 2 ) + i(x 1 y 2 + y 1 x 2 ) and it is denoted by z 1 z 2. For example: z 1 = 5+3i and z 2 = 2+4i then z 1 z 2 = (10-12) +i(20+6)=-2+26i.

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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 x 2 -y 2 +2xyi=3-4i ( (a+b) 2 =a 2 +b 2 +2ab) => x 2 –y 2 =3 ………………..(i) And 2xy=-4, i.e., xy=-2………..(ii) We have (x 2 +y 2 ) 2 = (x 2 -y 2 ) 2 +4x 2 y 2 3 2 +4(4)=9+16=25 x 2 +y 2 =+5 but as x 2 +y 2 =5………………..(iii) On adding (i) and (iii), we get 2x 2 =8 => x 2 =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 -2+i.

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Let x be a cube root of unity, then x 3 =1. => x 3 -1=0=> (x-1)(x 2 +x+1)=0 => either x-1=0 or x 2 +x+1=0 Either x=1or x=(-1+-3)/2 Thus, the three cube roots of unity are 1, (-1+i3)/2 and (-1-i3)/2.

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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 w 2. Further, to avoid any possible confusion, we shall take w= (-1+i3)/2 Thus, the three cube roots of unity are 1,w and w 2. Sum of the three cube roots of unity is zero. Thus 1+w+w 2 =0 Product of the cube roots of unity is one. i.e. 1.w.w 2 =w 3 =1 Either of the two non-real cube roots of unity is reciprocal of the other. Since w 3 =1, therefore, w.w 2 =1=> w and w 2 are reciprocals of each other.

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When we write the complex number in polar form, then z=r(cos φ + i sin φ ) Then according to De-Moivres Theorem z n = [r(cos φ + i sin φ )] n = r n (cos n φ + i sin n φ )

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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<0where d=b 2 -4ac Then roots are (–b+b 2 -4ac)/2a.

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Solve the equation 3x 2 +7=0 Here, the discriminant=0 2 -4x3x7=-84 Thereforex=(-0+-84)/2x3 =(+2(21)i)/6 => (+(21)i)/3

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