How do you divide # (7-i) / (3-i) # in trigonometric form?

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Answer 1 · 2018-08-11T11:56:34

In trigonometric form: #2.236(cos 0.18+i sin 0.18)#

# Z=(7-i)/(3-i)#

#Z=a+ib #. Modulus: #|Z|=sqrt (a^2+b^2)#;

Argument:#theta=tan^-1(b/a)# Trigonometrical form :

#Z =|Z|(costheta+isintheta)#

#Z_1= 7- i #.Modulus:#|Z_1|=sqrt(7^2+(-1)^2) #

#=sqrt 50 ~~ 7.07# Argument: #tan alpha= ((|-1|))/(|7|)#

#=1/7 , alpha =tan ^-1 (1/7) ~~ 0.142, Z # lies on fourth quadrant,

so #theta =2pi-alpha=2pi-0.142 ~~ 6.14#

# :. Z_1=7.07(cos 6.14+i sin 6.14) #,

#Z_2= 3- i #.Modulus:#|Z_2|=sqrt(3^2+(-1)^2) #

#=sqrt 10 ~~ 3.16# Argument: #tan alpha= ((|-1|))/(|3|)#

#=1/3 , alpha =tan ^-1 (1/3) ~~0.322, Z # lies on fourth quadrant,

so #theta =2pi-alpha=2pi-0.322 ~~ 5.96#

# :. Z_2=3.16(cos 5.96+i sin 5.96) #,

# Z=(7-i)/(3-i)#

# Z= (7.07(cos 6.14+i sin 6.14))/(3.16(cos 5.96+i sin 5.96)#

#Z=2.236(cos(6.14-5.96)+isin (6.14-5.96))# or

#Z=2.236(cos 0.18+i sin 0.18) =2.2+0.4 i#

In trigonometric form; #2.236(cos 0.18+i sin 0.18)# [Ans]