Question

Find the product of z1 and z2 where z1 = 2(cos 80° + i sin 80°), and z2 = 9(cos 110° + i sin 110°).

11(cos 190° + i sin 190°)

18(cos 8800° + i sin 8800°)

18(cos 190° + i sin 190°)

18(cos 80° + i sin 80°)

Answer 1

If

z1 = r1*( cos(theta1) + i*sin(theta1) )
z2 = r2*( cos(theta2) + i*sin(theta2) )


then

z1*z2 = (r1*r2) * ( cos(theta1 + theta2) + i*sin(theta1 + theta2) )

Answer 2

18(cos 190° + i sin 190°)
is this right?

Answer 3

correct

Answer 4

can u help me here?
Express the complex number in trigonometric form.

6 - 6i

six square root two times the quantity cosine of seven pi divided by four plus i times sine of seven pi divided by four

six square root two times the quantity cosine of five pi divided by four plus i times sine of five pi divided by four

six times the quantity cosine of seven pi divided by four plus i times sine of seven pi divided by four

six times the quantity cosine of five pi divided by four plus i times sine of five pi divided by four

Answer 5

6-6i is in the form a+bi

a = 6
b = -6

Answer 6

to convert to trigonometric form, you use these formulas

Answer 7

r = sqrt(a^2 + b^2)

theta = arctan(b/a)

Answer 8

theta=arctan (-6/6)
theta=pi/4
r=6
so it's either
\[6(\cos \frac{ 7\pi }{ 4 } + i \sin \frac{ 7\pi }{ 4 }) \]
\[6(\cos \frac{ 5\pi }{ 4 } + i \sin \frac{ 5\pi }{ 4 })\]
am i right?

Answer 9

(6,-6) is in quadrant IV, so you should have gotten -pi/4 which is coterminal to 2pi+(-pi/4) = 7pi/4

Answer 10

so it's

\[\large 6(\cos \frac{ 7\pi }{ 4 } + i \sin \frac{ 7\pi }{ 4 }) \]

Answer 11

ok thanks