Question

check answer @jim_thompson5910

Answer 1

hint:
\[z \cdot w = r\rho {e^{i\left( {\theta + \varphi } \right)}}\]

Answer 2

That's not the formula I used...but anyway...am I right or wrong?

Answer 3

hint:
\[{e^{i\theta }} = \cos \theta + i\sin \theta \]

Answer 4

I think you are right!

Answer 5

Thank you @Michele_Laino

Answer 6

thanks! @StudyGurl14

Answer 7

what's r?

Answer 8

r is the modulus of our complex number, for example r =3

Answer 9

sorry i'm having a dumb moment...what's the modulus?

Answer 10

the modulus r of a complex number z is given by the subsequent formula:
\[r = \sqrt {z \cdot {z^*}} \]

Answer 11

okay. what's the exponent? i can't read it

Answer 12

that is the Euler formula, namely:
\[{e^{i\theta }} = \cos \theta + i\sin \theta \]
so we have:
\[z = r{e^{i\theta }} = r\left( {\cos \theta + i\sin \theta } \right)\]

Answer 13

You are correct @StudyGurl14

I'm getting the same result

r1 = 3, theta1 = pi/3
r2 = 3, theta2 = 5pi/3

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

-------------------------------------------------------

z1*z2 = r1*r2*[cos(theta1+theta2)+i*sin(theta1+theta2)]
z1*z2 = 3*3*[cos(pi/3+5pi/3)+i*sin(pi/3+5pi/3)]
z1*z2 = 9*[cos(6pi/3)+i*sin(6pi/3)]
z1*z2 = 9*[cos(2pi)+i*sin(2pi)]
z1*z2 = 9*[1+i*0]
z1*z2 = 9*1
z1*z2 = 9

Answer 14

That's the formula I used^