How do you write the complex number in trigonometric form #-8-5sqrt3i#?

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Answer 1 · 2017-12-14T09:41:53

#z=sqrt(139)(cos47.3+isin47.3)#

If #z=a+bi# then #z=r(costheta+isintheta)#

#r=sqrt(a^2+b^2)#

#theta=tan^(-1)(b/a)#

#a=-8#
#b=-5sqrt(3)#

#r=sqrt((-8)^2+(-5sqrt(3))^2)=sqrt(139)#

#theta=tan^(-1)((-5sqrt(3))/(-8))~~=47.3^circ#

#z=sqrt(139)(cos47.3+isin47.3)#