We know that the solution of the quadratic equation
ax2 + bx + c = 0, a≠0
are given by quadratic formula
x = -b ± √(b2-4ac)/2a
Now when b2-4ac < 0 the equation cannot be solved in real number system. For this solution, we need the set of complex numbers.
i= √(-1)
The set of complex numbers denoted by ℂ is defined as:
ℂ = {a+bi∶a,b ∈ R,i = √(-1)}
The notation a + bi is called writing complex number in standard form
a is called the real part of complex number z= a + bi (also written as a = Re z) and b is called the imaginary part of complex number z = a+bi (also written as b = Im z ).
Examples of complex numbers can be given as 2+4i or √(-9 )=9i etc.
Two complex numbers z1= a1+b1 i and z2= a2+ b2 i are equal if their real and imaginary parts are equal, which means a1=a2 and b1= b2.
Every real number is a complex number with its imaginary part 0. Thus a = a + 0i
So R ⊂ ℂ
The complex number z̅= a - bi is called the conjugate of a complex number z = a + bi .
The absolute value of complex number z is defined as |z|= √(a2+b2)
Two complex numbers can be added, subtracted, multiplied and divided resulting in new complex numbers.
If z1= a1 + b1 i and z2 = a2 + b2 i then
z1 + z2 = (a1 + a2 ) + (b1 + b2 )i
z1 - z2 = (a1 - a2 )+(b_1-b_2 )i
z1 z2 = (a1 a2 - b1 b2 )+(a1 b2 + a2 b1 )i
z1/z2 = a1 + b1 i/a2 + b2 i = (a1 + b1 i)( a2 - b2 i) )/(a2 + b2 i)(a2- b2 i) = (a1 a2+ b1 b2)/(a22 + b22 ) - (a1 b2 + a2 b1)/(a22 + b22 ) i
Then,
1/z= 1/(a+bi)= (a-bi)/((a+bi)(a-bi)) = (a-bi)/(a2+b2 )= a/(a2+b2 ) - b/(a2+b2 ) i
For simplifying complex numbers we must remember the various powers of imaginary number I, especially while multiplying imaginary numbers:
Let z1 = 3 - 2i and z2 = -2 + 3i then
Re z1 = 3, Im z1 = -2 and
Re z2 = -2, Im z2 = 3
|z1 | = √(32 + (-2)2 ) = √13
|z2 | = √((-2)2+ 32 )= √13
z1 + z2 = (3-2)+(-2+3)i = 1+i
z1 -z2 = (3 -( -2)) + (-2 -3)i = 5-5i
z1 z2 = (3*(-2)-(-2*3))+(3*3+(-2*-2))i = 13i
z1/z2 = 3-2i/-2+ 3i = 3*(-2) + (-2*3)/4+9 - 3*3 +(-2 * -2))/4+9 i = -12/13 + 13/13 i= -12/13 + i
1/z1 = 1/3-2i = 3/13 + 2/13 i
1/z2 = 1/-2+3i = -2/13 - 3/13 i
z̅1 =3 +2i
z̅2 = -2 -3i
z̅1 + z̅2 = (3 - 2)+(2 - 3)i = 1-i
Hence verified:
z̅1 . z̅2 = (3*(-2)-(-2*3)) + (3*-3+(2*-2))i = -13i
z̅1̅z̅2̅ = 1̅3̅i = - 13i
Hence verified: z̅1̅z̅2̅ = z̅1 . z̅2
z̅1.z1 = (3-2i)*(3+2i)=(9+4)+(-2*3+3*2)i=13=(√13)2 = |z1 |2
Hence verified: z̅.z = |z|2
z̅1 + z1 = (3-2i)+(3+2i)= 6 = 2*3 = 2Re z1
Hence verified: z+z̅ = 2 Re z
z1 - z1̅ =(3-2i) - (3+2i) = -4i = 2i(-2) = 2i Im z1
Hence verified: z-z̅ = 2i Im z
Hence verified: Re z ≤ |z|
z̿1 = 3̿-̿2̿i̿ = (3+2i) ̅ = 3-2i = z1
Hence verified: z̿ = z
Complex numbers have vast applications in the field of electrical engineering, study of electrical circuits, trigonometry, calculus, quantum mechanics, and study of waves such as electricity, light, and sound. Or simply in finding square root of negative numbers.
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