Arithmetic operations over Complex Numbers

In general, arithmetic of complex numbers is same as arithmetic of real numbers, we just handle real and imaginary parts separately and remember that `i^2=-1` .

Sum of complex numbers `z_1=a+bi` and `z_2=c+di` is complex number `z_1+z_2=(a+bi)+(c+di)=(a+c)+(b+d)i`.

For example, `(3+4i)+(2+7i)=(3+2)+(4+7)i=5+11i`.

Difference of complex numbers `z_1=a+bi` and `z_2=c+di` is complex number `z_1-z_2=(a+bi)-(c+di)=(a-c)+(b-d)i`.

For example, `(3+4i)-(2+7i)=(3-2)+(4-7)i=1+(-3)i=1-3i`.

Product of complex numbers `z_1=a+bi` and `z_2=c+di` is complex number `z_1z_2=(a+bi)(c+di)=ac+adi+bci+bidi=ac+(ad+bc)i+bdi^2=ac+(ad+bc)i+bd(-1)=(ac-bd)+(ad+bc)i`.

For example, `(3+4i)(2+7i)=(3*2-4*7)+(3*7+4*2)i=-22+29i`.

To find division of two numbers `z_1=a+bi` and `z_2=c+di!=0` we need to multiple numerator and denominator by the complex conjugate of denominator.

`(z_1)/(z_2)=(a+bi)/(c+di)=((a+bi)(c-di))/((c+di)(c-di))=((ac+bd)+(bc-ad)i)/(c^2+d^2)=(ac+bd)/(c^2+d^2)+(bc-ad)/(c^2+d^2)i`.

For example, `(3-4i)/(2+7i)=(3*2+(-4)*7)/(2^2+7^2)+((-4)*2-3*7)/(2^2+7^2)i=-22/53-29/53i`.

There is no need to remember these formulas. The main thing is to now how to derive them.

Example. Find `(1+2i)i-(3+2i)/(1-i)`.

First we perform multiplication: `(1+2i)i=i+2i^2=i+2(-1)=-2+i`.

Now, we perform division: `(3+2i)/(1-i)=((3+2i)(1+i))/((1-i)(1+i))=(3+3i+2i+2i^2)/(1+i-i-i^2)=(3+5i-2)/(1+1)=(1+5i)/2=1/2+5/2i`.

Finally, `(1+2i)i-(3+2i)/(1-i)=(-2+i)-(1/2+5/2 i)=-2-1/2+i-5/2i=-5/2-3/2i`.

To find power of complex number, we use sum of squares and sum of cubes formula and perform same arithmetic as with real numbers.

It is also convenient to use general rule for finding power of `i`: since `i^1=i`, `i^2=-1`, `i^3=-i`, `i^4=1` then `i^(4n+1)=i, i^(4n+2)=-1, i^(4n+3)=-i, i^(4n+4)=1`, `n in N`.

Example. Find `(1+2i)^3`.

`(1+2i)^3=1^3+3*1^2*2i+3*1*(2i)^2+(2i)^3=1+6i+3*1*4*i^2+2^3i^3=`

`=1+6i-12-8i=-11-2i`.