How do you divide #( 4i+4) / (6i +5 )# in trigonometric form?

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Answer 1 · 2018-02-27T12:08:15

In trigonometric form: #0.725(cos 0.091-isin 0.091)#

# (4+4i)/(5+6i) ;Z=a+ib #. Modulus: #|Z|=sqrt (a^2+b^2)#;

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

#Z =|Z|(costheta+isintheta)# #Z_1= 4+4 i #

Modulus:#|Z_1|=sqrt(4^2+4^2)~~ 5.66 #

Argument: #tan alpha= (|4|)/(|4|):. alpha = tan^-1 (1)=0.785#

#Z_1# lies on first quadrant, so #theta =alpha ~~ 0.785#

# :. Z_1=5.66(cos 0.785+isin 0.785) #

#Z_2= 5 + 6i #. Modulus:#|Z|=sqrt(5^2+6^2) #

#=sqrt 61 ~~ 7.81# Argument: #tan alpha= (|6|)/(|5|)#

#=6/5 :.alpha =tan^-1 (1.2) ~~ 0.0876 ; Z_2# lies on first

quadrant.#:. theta=alpha ~~0.876#

# :. Z_2=7.81(cos 0.876+isin 0.876) :. (4+4i)/(5+6i) = #

# Z= (5.66(cos0.785+isin 0.785))/(7.81(cos 0.876+isin 0.876)#

#Z=0.725(cos(0.785-0.876)+isin (0.785-0.876))# or

#Z=0.725(cos (-0.091)+isin (-0.091)) # or

#Z=0.725(cos (0.091)-isin (0.091))=(44/61-4/61 i )#

In trigonometric form: #0.725(cos 0.091-isin 0.091)# [Ans]