Computung Complex Numbers.?

Addition is easy; you just add the real parts together and the imaginary parts together.

a+bi + c+di = (a+c)+(b+d)i

eg. 2+3i + 4+7i = 6+10i

Subtraction is the same; just subtract the real part from the real part and imaginary from imaginary.

a+bi – c+di = (a-c)+(b-d)i

eg 2+3i – 4+7i = (-2)+(-4)i which can also be written -2-4i or -(2+4i)

Multiplication is just as you would multiply out brackets normally although the i makes things look more confusing. Just treat the i like everything else and it will all work out:

(a+bi)(c+di)=(ac + adi + bci + bdi^2)

remember i^2 = -1.

so = (ac + adi + bci + (-1)bd)

= (ac + adi + bci – bd)

Then collect the terms: (ac-bd)+(ad+bc)i

eg. (2+3i)(4+2i)=(8+12i+4i+6i^2)

=(8+12i+4i-6)

=14+16i

Division is the hardest one.

If you have (a+bi)/(c+di), first multiply top and bottom by what we call the complex conjugate of the denominator (the bottom of the division). The complex conjugate of c+di is c-di, and vice versa. eg complex conjugate of 2+3i is 2-3i.

We are allowed to do this because multiplying the top and bottom of a fraction or a division sum by the same thing doesn’t change the fraction. We do it because, as you will see, multiplying a complex number by its conjugate removes the imaginary numbers.

So, for (a+bi)/(c+di), multiplying top and bottom by (c-di) gives

[(a+bi)(c-di)]/[(c+di)(c-di)]

Multiply out:

[ac + bci – adi -bdi^2]/[c^2 + cdi – cdi -(d^2)(i^2)]

Remembering i^2 = -1, and cancelling, this becomes:

[ac + bci – adi + bd]/[c^2 + d^2]

= [(ac+bd)+(bc-ad)i)]/[c^2 + d^2]

which is the answer.

eg. (2+3i)/(1+2i). multiply by complex conjugate of denominator:

= [(2+3i)(1-2i)]/[(1+2i)(1-2i)]

multiply out:

=[2 + 3i – 4i -6i^2]/[1^2 + 2i – 2i -(4^2)(i^2)]

=[2 + 3i – 4i +6]/[1 + (4^2)]

=(8-i)/(1+16)

=(8/17)-(1/17)i

But it would be easier to answer if you posed a specific problem, and we attempted to show you how to answer it.