Find the quotient and write the result in a+bi form? #(16(cos240^\circ+isin240^\circ)) /(8(cos90^\circ+isin90^\circ))#

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#(16(cos240^\circ+isin240^\circ)) /(8(cos90^\circ+isin90^\circ))#

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Answer 1 · 2017-12-12T18:13:24

#-sqrt(3)+i#

They're of the form #r*(cos(theta)+isin(theta))#, where #r# is the modulus and #theta# is the argument.

To divide complex numbers in trig form you divide their moduli and subtract their arguments: #16/8 = 2# and #240^circ -90^circ = 150^circ#:

#(16(cos240^\circ+isin240^\circ)) /(8(cos90^\circ+isin90^\circ))=#

#2(cos(150^circ)+isin(150^circ))=#

#2(-sqrt(3)/2+1/2i)=#

#-sqrt(3)+i#