Calculate [cos(3pi/4) + isin(3pi/4)]*[cos(-5pi/6) + isin(-5pi/6)]

We'll apply the rule of multiplying 2 complex number, put
in polar form:

[cos (a1) + i*sin (a1)]*[cos (a2) + i*sin
(a2)] = [cos (a1+a2) + i*sin (a1+a2)]

We'll also apply
Moivre's rule:

[cos (a1) + i*sin (a1)]^n = [cos (n*a1) +
i*sin (n*a1)]

We'll apply Movre's rule to the first pair
of brackets:

[cos(pi/4) + isin(pi/4)]^3 = [cos(3pi/4) +
isin(3pi/4)] (1)

We'll apply Movre's rule to the next
brackets:

[cos(pi/6) + isin(pi/6)]^-5 = [cos(-5pi/6) +
isin(-5pi/6)] (2)

We'll multiply (1) and
(2):

[cos(3pi/4) + isin(3pi/4)]*[cos(-5pi/6) +
isin(-5pi/6)] = [cos(3pi/4 - 5pi/6) + isin(3pi/4 -
5pi/6)]

[cos(3pi/4 - 5pi/6) + isin(3pi/4 - 5pi/6)] =
{cos[(9pi-10pi)/12] + isin[(9pi-10pi)/12]}

[cos(3pi/4 -
5pi/6) + isin(3pi/4 - 5pi/6)] = cos (-pi/12) +
isin(-pi/12)

Since cosine function is even and sine
function is odd, we'll get:

[cos(3pi/4) +
isin(3pi/4)]*[cos(-5pi/6) + isin(-5pi/6)] = cos (pi/12) -
isin(pi/12)