# 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**.

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`.

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))`.