Find the cube roots of 125(cos 288° + i sin 288°).
we have :
and we know that cosx+isinx can be writeen as e^i(x) , thus :
=5*e^(i*288*1/3) = 5*e^(i96)
hope it helps
let r = 125(cos(288Â°) + i sin(288Â°)). First, you can rewrite this complex number as
r = 125 * exp(i 288Â°)
This is called the polar representation with the complex exponential function. Now you want to find a number a that satisfies a^3 = r = 125 * exp(i 288Â°).
You know that by multiplying two complex numbers x and y, their radii multiply and their polar angles add. That means the number a you are looking for has radius 125^(1/3) = 5. For the polar angle b of the complex number a it holds 3b = 288Â°. It looks as this equation has only one solution but it has three, because 288Â° means the same on the complex plane as 288Â°+360Â° and 288Â°+720Â°. So, the polar angles of a can be 288Â°/3=96Â°, (288Â°+360Â°)/3 = 216Â° and (288Â° + 720Â°)/3 = 336Â°. Therefore,
all cube roots of your number are
5(cos 96Â° + i sin 96Â°),
5(cos 216Â° + i sin 216Â°) and
5(cos 336Â° + i sin 336Â°).
Â³â125 = 5 and 288/3 = 96 and 360/3 = 120, so
5(cos 96 + i sin 96)
5(cos 216 + i sin 216)
5(cos 336 + i sin 336)
5(cos( (288Â° + k360Â°)/3) + i sin ( (288Â° + k360Â°)/3 ) ) for k=0,1,2
z_0 = 5(cos 96Â° + i sin 96Â°)
z_1 = 5(cos 216Â° + i sin 216Â°)
z_2 = 5(cos 336Â° + i sin 336Â°)