How do you write the complex number in trigonometric form #5-12i#?

1 Answer
Jun 6, 2017

In trigonometric form : #13(cos292.62+isin292.62)#

Explanation:

#Z=a+ib #. Modulus: #|Z|=sqrt (a^2+b^2)#; Argument:#theta=tan^-1(b/a)# Trigonometrical form : #Z =|Z|(costheta+isintheta)#

#Z=5-12i #. Modulus #|Z|=sqrt(5^2+(-12)^2) =sqrt(25+144)=sqrt169=13#

Argument: #tan alpha = 12/5= 2.4 #. Z lies on fourth quadrant, #alpha =tan^-1(2.4) = 67.38^0 :. theta = 360-67.38=292.62^0 :. Z=13(cos292.62+isin292.62)#

In trigonometric form expressed as #13(cos292.62+isin292.62)# [Ans]