How do you express the complex number in trigonometric form: #-7-5i#?

2 Answers
Monzur R.
Aug 23, 2017

#-7-5i = sqrt74(cos2.52-isin2.52)#

Explanation:

Any #a+bi# can be rewritten in mod-arg form as #r(cosvartheta +isinvartheta)# where #r = sqrt(a^2+b^2)# and #vartheta = arctan (b/a)#

#|-7-5i| = sqrt(5^2+7^2)=sqrt74#

#"arg"(-7-5i) = arctan ((-5)/(-7)) = -2.52#

#therefore -5-7i -= sqrt74 (cos(-2.52)+isin (-2.52)) = sqrt74(cos2.52-isin2.52)#

Jim G.
Aug 23, 2017

#sqrt74(cos(2.52)-isin(2.52))#

Explanation:

#"to convert from "color(blue)"complex to trig. form"#

#"that is "x+yitor(costheta+isintheta)" using"#

#•color(white)(x)r=sqrt(x^2+y^2)#

#•color(white)(x)theta=tan^-1(y/x)color(white)(x)-pi < theta<=pi#

#"here "x=-7" and "y=-5#

#rArrr=sqrt((-7)^2+(-5)^2)=sqrt74#

#-7-5i" is in the third quadrant so we must ensure "theta#
#"is in the third quadrant"#

#theta=tan^-1(5/7)=0.62larrcolor(blue)" related acute angle"#

#rArrtheta=-pi+0.62=-2.52larrcolor(blue)" in third quadrant"#

#rArr-7-5itosqrt74(cos(-2.52)+isin(-2.52))#

#=sqrt74(cos(2.52)-isin(2.52))#

Related questions
See all questions in Trigonometric Form of Complex Numbers
Impact of this question
2680 views around the world
You can reuse this answer
Creative Commons License