# Operations over Complex Numbers in Trigonometric Form. De Moivre's Formula

Suppose that we are given two complex numbers in trigonometric form: z_1=r_1(cos(phi_1)+isin(phi_1)) and z_2=r_2(cos(phi_2)+isin(phi_2)).

Then their product is color(red)(z_1z_2=r_1r_2(cos(phi_1+phi_2)+isin(phi_1+phi_2))) and their quotient is color (blue)(z_1/z_2=r_1/r_2(cos(phi_1-phi_2)+isin(phi_1-phi_2))).

Example 1. Find z_1z_2 if z_1=4(cos(pi/6)+isin(pi/6)) and z_2=1/10(cos(-(2pi)/3)+isin(-(2pi)/3)).

We use formula for product: z_1z_2=4* 1/10(cos(pi/6+(-(2pi)/3))+isin(pi/6+(-(2pi)/3)))=

=2/5(cos(-pi/2)+isin(-pi/2))=-2/5i.

Example 2. Find (32(cos((2pi)/3)+isin((2pi)/3)))/(8i).

We first need to find trigonometric form of complex number 8i: 8i=8(cos(pi/2)+isin(pi/2)).

Now we use formula for quotient: (32(cos((2pi)/3)+isin((2pi)/3)))/(8i)=(32(cos((2pi)/3)+isin((2pi)/3)))/(8(cos(pi/2)+isin(pi/2)))=32/8(cos((2pi)/3-pi/2)+isin((2pi)/3-pi/2))=

=4(cos(pi/6)+isin(pi/6))=4(sqrt(3)/2+i 1/2)=2sqrt(3)+2i.

If we want to raise complex number z=r(cos(phi)+isin(phi)) to the natural power n, then we use following formula: color(green)(z^n=r^n(cos(n phi)+isin(n phi))).

When r=1, we obtain De Moivre's Formula (or Theorem) color(red)((cos(phi)+isin(phi))^n=cos(n phi)+isin(n phi)).

Example 3. Find z^7 if z=2(cos(-(3pi)/4)+isin(-(3pi)/4)).

Using above formula we have that z^7=2^7(cos(7*(-(3pi)/4))+isin(7*(-(3pi)/4)))=128(cos(-(21pi)/4)+isin(-(21pi)/4))=

=128(cos((3pi)/4)+isin((3pi)/4))=128(-sqrt(2)/2+sqrt(2)/2)=-64sqrt(2)+64sqrt(2)i.

To find n-th natural root of complex number, following formula is used: color(blue)(root(n)z=root(n)r(cos((phi+2pik)/n)+isin((phi+2pik)/n))) where k=0,1,2,...,n-1.

Example 4. Solve equation z^2+4=0.

We have that z^2=-4 or z=sqrt(-4).

We need trigonometric form of -4: -4=4(cos(pi)+isin(pi)) .

Therefore, sqrt(-4)=(-4)^(1/2)=sqrt(4)(cos((pi+2pik)/2)+isin((pi+2pik)/2)), k=0,1.

If k=0 then sqrt(-4)=sqrt(4)(cos((pi+2pi*0)/2)+isin((pi+2pi*0)/2))=

=2(cos(pi/2)+isin(pi/2))=2i.

If k=1 then sqrt(-4)=sqrt(4)(cos((pi+2pi*1)/2)+isin((pi+2pi*1)/2))=

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

Therefore, equation z^2+4=0 has 2 roots: 2i and -2i.

Example 5. When we were asked to solve equation x^3=64, we just wrote that x=4. It is not quite true. Since this is equation of third degree then it should have 3 roots. One is real, but other two are complex. Let's find them.

We have that x=root(3)64.

We need trigonometric form of 64: 64=64(cos(0)+isin(0)).

So, root(3)64=root(3)64(cos((0+2pik)/3)+isin((0+2pik)/3)) where k=0,1,2.

If k=0 then root(3)64=root(3)64(cos((0+2pi*0)/3)+isin((0+2pi*0)/3))=

=4(cos(0)+isin(0))=4(1+i*0)=4 (this is real number solution).

If k=1 then root(3)64=root(3)64(cos((0+2pi*1)/3)+isin((0+2pi*1)/3))=

=4(cos((2pi)/3)+isin((2pi)/3))=4(-1/2+i(sqrt(3)/2))=-2+2sqrt(3)i.

If k=2 then root(3)64=root(3)64(cos((0+2pi*2)/3)+isin((0+2pi*2)/3))=

=4(cos((4pi)/3)+isin((4pi)/3))=4(-1/2+i*(-sqrt(3)/2))=-2-2sqrt(3).

Therefore, there are 3 solutions to the equation x^3=64: x=4, x=-2+2sqrt(3)i and x=-2-2sqrt(3).