Cos 2pi 3

Z = 5 (in 2pi / 3 + in sin 2pi / 3) writes complex numbers in regular form.

r = 5

Theta = 2 pi / 3

X = rkos (theta) = 5kos (2 pi / 3) = 5 (1/2) = 5/2

y = r sin (theta) = 5 sins (2pi / 3) = 5 sqrt (3) / 2

5/2 + [5 squared (3) / 2] i

Z = 5 (because sin in 2 € / 3 + 2 € / 3) = 5 (1/2 + ˆš3 / 2 i) = 5/2 + 5ˆ ˆš 3/2 i i    Answer

Cos 2pi 3

Cos 2pi 3

Z = 5 (cos 2pi / 3 + i sin 2pi / 3) writes complex numbers in a regular format. ۔

r = 5.

Theta = 2pi / 3.

x = r cos (theta) = 5 cos (2pi / 3) = 5 (1/2) = 5/2.

y = r sin (theta) = 5 sin (2pi / 3) = 5 sqrt (3) / 2

5/2 + [5sqrt (3) / 2] i.

Z = 5 (cos 2 € / 3 + i sin 2 € / 3) = 5 (1/2 + šš3 / 2 i) = 5/2 + 5º 3/2 i  NS Answer

Cos 2pi 3

Cos 2pi 3

Z = 5 (cos 2pi / 3 + i sin 2pi / 3) writes complex numbers in a recular form. 3

Z = 5 (cos 2 € / 3 + i sin 2 € / 3) = 5 (1/2 + šš3 / 2 i) = 5/2 + 5º 3/2 i  Answer

Cos 2pi 3