OpenStudy (anonymous):

Convert the polar representation of this complex number into its standard form: z=6(cos300+isin300) [300 is in degrees] HELP PLEASE

4 years ago
OpenStudy (anonymous):

@johnweldon1993

4 years ago
OpenStudy (akashdeepdeb):

Z = a + ib Z = 6(cos300+isin300) a + ib = 6(cos300+isin300) a + ib = 6cos 300 + i . 6sin 300 = 6 cos (360 - 60) + i. 6sin (360 - 60) = 6 cos (-60) + i . 6 sin (-60) z = 6 cos 60 - i . 6 sin (60) Getting this? :)

4 years ago
OpenStudy (anonymous):

Okay i got that but what do u do to put it into complex form?

4 years ago
OpenStudy (anonymous):

I mean standard form

4 years ago
OpenStudy (anonymous):

4 years ago
OpenStudy (akashdeepdeb):

That *is* the standard form that I solved there. Look, standard form of a complex number (z) = A + iB Here you are given in polar co-ordinates. Just put it in terms of A + iB [Put it in that form] NOTE: Find cos 60 and sin 60. You'll have you answer. Getting this?

4 years ago
OpenStudy (anonymous):

So D is the correct answer right?

4 years ago
OpenStudy (anonymous):

@AkashdeepDeb

4 years ago
OpenStudy (johnweldon1993):

Great job @AkashdeepDeb and yes caseyf D would be correct here, just by the fact that cos(60) = 1/2 and sin(60) = √3/2

4 years ago
OpenStudy (anonymous):

Okay thank you!

4 years ago