What is the trigonometric form of # (-10+3i) #?

1 Answer
SagarStudy
Dec 30, 2015

#(-10+3i)=sqrt109(Cos(tan^-1(-3/10))+isin(tan^-1(-3/10)))#

Explanation:

If #(a+ib)# is a complex number, #u# is its magnitude and #alpha# is its angle then #(a+ib)# in trigonometric form is written as #u(cosalpha+isinalpha)#.
Magnitude of a complex number #(a+ib)# is given by#sqrt(a^2+b^2)# and its angle is given by #tan^-1(b/a)#

Let #r# be the magnitude of #(-10+3i)# and #theta# be its angle.
Magnitude of #(-10+3i)=sqrt((-10)^2+3^2)=sqrt(100+9)=sqrt109=r#
Angle of #(-10+3i)=Tan^-1(3/-10)=tan^-1(-3/10)=theta#

#implies (-10+3i)=r(Costheta+isintheta)#
#implies (-10+3i)=sqrt109(Cos(tan^-1(-3/10))+isin(tan^-1(-3/10)))#

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