How do I divide #6(cos^circ 60+i\ sin60^circ)# by #3(cos^circ 90+i\ sin90^circ)#?

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Answer 1 · 2015-11-08T20:42:19

#2cis(-30^@)=sqrt3-i#

The easiest way to divide complex numbers is to first convert to polar form and then you may divide modulus separately and subtract the angles.

Let #z_1=r_1cistheta_1=6cis60^@=6cos60^@+6isin60^@=6/_60^@=6/_pi/3#

#z_2=r_2cistheta_2=3cis90^@=3cos90^@+3isin90^@=3/_90^@=3/_pi/2#

#therefore z_1/z_2=r_1/r_2cis(theta_1-theta_2)#

#=6/3cis(60^@-90^@)#

#=2cis(-30^@)=2/_-30^@#

#=2(cos30^@-isin30^@)#

#=sqrt3-i#