Multiplication of complex numbersMultiplication of complex numbers
The complex number is written in the form z = a+ bi.
We need to determine the result of the product of two complex numbers.
Let z1= a+ bi
and z2 = c+di
Now we need to calculate the product of z1*z2.
==> z1*z2= (a+bi)(c+di)
We will expand the brackets.
==> z1*z2= (a*c + bc*i + ad*i + bd*i^2)
Now we know that i^2 = -1
==> z1*z2= ac + bc*i + ad*i - bd )
Now we will combine real terms and complex terms together.
==> z1*z2= (ac-bd) + (bc+ad)*i
Now I will give an example:
Find the product of ( 2-3i) * ( 3+ 5i)
Let us expand.
==> (2*3 - 3*3*i + 2*5i -3i*5i
==> (6 - 9i + 10i - 15i^2)
i^2 = -1
==> (6 + i -15) = (-9 + i)
Since in a former post, I've explained the multiplication of complex numbers written in algebraic form, now I'll show how to multiply 2 complelx numbers put in trigonometric form.
z1 = cos a + isin a and z2 = cos b + isin b
The real part of z1 = Re(z1) = cos a
The imaginary part of z1 = Im(z1) = sin a
The real part of z2 = Re(z2) = cos b
The imaginary part of z2 = Im(z2) = sin b
We'll multiply the numbers z1*z2 = (cos a + isin a )(cos b + isin b)
We'll apply FOIL:
z1*z2 = cos a*cosb + i(sina*cosb + sinb*cos a) + i^2*sin a*sin b
Rule number one: i^2 = -1
z1*z2 = cos a*cosb + i(sina*cosb + sinb*cos a) - sin a*sin b
Rule number 2: We'll combine real parts and imaginary parts:
z1*z2 = (cos a*cosb - sin a*sin b) + i(sina*cosb + sinb*cos a)
We recognize the formula of cosine of the sum of 2 angle, to the real part:
cos a*cosb - sin a*sin b = cos (a+b)
We recognize the formula of sine of the sum of 2 angle, to the imaginary part:
sina*cosb + sinb*cos a = sin (a+b)
z1*z2 = cos (a+b) + i*sin (a+b)