Geometric Representation of Complex Numbers. Trigonometric Form of Complex Numbers

Complex number z=a+bi on coordinate pllane xOy is represented by point M with coordinates (a,b). x-axis is called real axis and y-axis is called imaginary axis.geometric representation of complex number

Real numbers are represented by points of real axis and purely imaginary numbers are represented by points of imaginary axis.

On the figure there are depicted 8 complex numbers: `z_1=2+i`, `z_2=3`, `z_3=2i`, `z_4=-1+i`, `z_5=-2.5`, `z_6=-1-i`, `z_7=-3i`, `z_8=3-2i`.

Corresponding points are (2,1), (3,0), (0,2), (-1,1), (-2.5,0), (-1,-1), (0,-3), (3,-2).

Note that points that corresponds to complex conjugates are symmetric with respect to x-axis (`z_4` and `z_6`).

We can also represent complex number in polar coordinates where arguments parameters are r and `phi`.

polar representation of complex number

From figure it is clear that `r=|OM|=sqrt(a^2+b^2)` (by Pythagorean theorem) is a an absolute value of z=a+bi. Polar angle `phi` is called argument of complex number that is represented by this point. Argument of complex number is defined ambiguously: if `phi` is argument of complex number z then `phi+2pik` is also argument of z when k is integer. For uniqueness we choose argument in interval `-pi<phi<=pi` and denote it by arg z; such value of argument is called main.

Definition. Trigonometric form of complex number z=a+bi is the following form: `z=r(cos(phi)+isin(phi))`, where `r=sqrt(a^2+b^2)` is absolute value and `phi` is argument of z.

Argument `phi` is connected with a and b by following formulas: `cos(phi)=a/sqrt(a^2+b^2)` and `sin(phi)=b/(sqrt(a^2+b^2))`.

Example 1. Find trigonometric form of `z=-2sqrt(3)+2i`.

First, find absolute value of `r=sqrt((-2sqrt(3))^2+2^2)=sqrt(12+4)=sqrt(16)=4`.

Now, we have that `cos(phi)=a/sqrt(a^2+b^2)=(-2sqrt(3))/4=-sqrt(3)/2` and `sin(phi)=b/sqrt(a^2+b^2)=2/4=1/2`. This means that `phi` lies in second quadrant and therefore `phi=(5pi)/6`.

So, `-2sqrt(3)+2i=4(cos((5pi)/6)+isin((5pi)/6))`.

Example 2. Find trigonometric form of `z=-3i`.

First, find absolute value of `r=sqrt(0^2+(-3)^2)=sqrt(0+9)=sqrt(9)=3`.

Now, we have that `cos(phi)=a/sqrt(a^2+b^2)=0/3=0` and `sin(phi)=b/sqrt(a^2+b^2)=(-3)/3=-1`. This means that`phi=-pi/2`.

So, `-3i=3(cos(-pi/2)+isin(-pi/2))`.

Example 3. Find trigonometric form of `z=1-sqrt(5)`.

First, find absolute value of `r=sqrt((1-sqrt(5))^2+0^2)=sqrt((1-sqrt(5))^2)=sqrt(5)-1`, (remember that absolute value should be positive).

Now, we have that `cos(phi)=a/sqrt(a^2+b^2)=(1-sqrt(5))/(sqrt(5)-1)=-1` and `sin(phi)=b/sqrt(a^2+b^2)=0/(sqrt(5)-1)=0`. This means that `phi=pi`.

So, `1-sqrt(5)=(sqrt(5)-1)(cos(pi)+isin(pi))`.