How do you divide # (1-2i) / (6+i) # in trigonometric form?

Question

1328 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-30T12:42:16

In trigonometric form: #0.368(cos 5.011+isin 5.011)#

# Z=(1-2i)/(6+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= 1-2 i #.Modulus:#|Z|=sqrt(1^2+(-2)^2) #

#=sqrt 5 ~~ 2.236# Argument: #tan alpha= (|-2|)/(|1|)#

#=2 #. #alpha =tan^-1 2 = 1.107; Z_1# lies on fourth quadrant, so

#theta =2pi-alpha=2pi-1.107 ~~ 5.176#

# :. Z_1=2.236(cos5.176+isin 5.176) #,

#Z_2= 6 + i #.Modulus:#|Z|=sqrt(6^2+1^2) #

#=sqrt 37 ~~ 6.083# Argument: #tan alpha= (|1|)/(|6|)#

#=1/6 :.alpha =tan^-1 (1/6) = 0.165 ; Z_2# lies on first quadrant,

#:. theta=alpha ~~0.165# # :. Z_2=6.083(cos 0.165+isin 0.165) #

# Z=(1-2i)/(6+i)#

# Z= (2.236(cos5.176+isin 5.176))/(6.083(cos 0.165+isin 0.165)#

#Z=0.368(cos(5.176-0.165)+isin (5.176-0.165))# or

#Z=0.368(cos 5.011+isin 5.011) =4/37-13/37i#

In trigonometric form: #0.368(cos 5.011+isin 5.011)# [Ans]