Multiplication of complex numbersMultiplication of complex numbers

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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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)

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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)

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