# 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 thatphi=-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)).