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

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

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

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

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

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

$=sqrt3-i$

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