Value of (cosπ/4+isinπ/4) ^4 is………

Method 2:
For imaginary number’s multiplication we have,
(a+ib)(c+id) = (ac – bd)+ i(ad + bc) ————– Equation 1

In the question, a = c = cos(pi/4) and b = d = sin(pi/4);
ac = (cos(pi/4))^2, ————– Equation 2
bd = (sin(pi/4))^2, ————– Equation 3
ad = bc = cos(pi/4)sin(pi/4) ————– Equation 4

(cos(pi/4) + sin(pi/4))^4 = [(cos(pi/4) + sin(pi/4))^2]^2
For convenience let (cos(pi/4) + sin(pi/4))^2 = t,
Question now is t^2 = ?
So,
t = (cos(pi/4) + sin(pi/4))^2 = [cos(pi/4) + sin(pi/4)][cos(pi/4) + sin(pi/4)]
Putting Equation 2,3,and 4 in 1, t will become;
t = [cos^2(pi/4) – sin^2(pi/4)] + i[cos(pi/4)sin(pi/4) + cos(pi/4)sin(pi/4)]
= [cos^2(pi/4) – sin^2(pi/4)] + i[2sin(pi/4)cos(pi/4)]

Now, cos^2X – sin^2X = cos2X, 2sinXcosX = sin2X, putting these values above;
t = cos(2pi/4) + i[sin(2pi/4)] = cos(pi/2) + i[sin(pi/2)] = 0 + i = i

Note: cos(pi/2) = 0, sin(pi/2) = 1, i^2 = -1

t = cos(pi/2) + i[sin(pi/2)] = 0 + i = i

Hence, t^2 = (i)^2 = -1