How do you divide #( 6i+5) / ( 7 i -4 )# in trigonometric form?

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Answer 1 · 2018-02-14T13:10:31

In trigonometric form: #0.969(cos 1.214-isin 1.214)#

# (5+6i)/(-4+7i)# #Z=a+ib #. Modulus: #|Z|=sqrt (a^2+b^2)#;

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

#Z =|Z|(costheta+isintheta) ; Z= 5+6 i #.

Modulus:#|Z|=sqrt(5^2+6^2)~~ 7.81 #

Argument: #tan alpha= (|6|)/(|5|):. alpha = tan^-1 (1.2)=0.876#

#Z_1# lies on first quadrant, so #theta =alpha ~~ 0.876#

# :. Z_1=7.81(cos 0.876+isin 0.876) #

#Z_2= -4 + 7i #. Modulus:#|Z|=sqrt(4^2+7^2) #

#=sqrt 65 ~~ 8.062# Argument: #tan alpha= (|7|)/(|-4|)#

#=7/4 :.alpha =tan^-1 (7/4) = 1.052 ; Z_2# lies on second

quadrant.#:. theta=pi-alpha ~~2.09#

# :. Z_2=8.062(cos 2.09+isin 2.09) :. (5+6i)/(-4+7i) =#

# Z= (7.81(cos0.876+isin 0.876))/(8.062(cos 2.09+isin 2.09)#

#Z=0.969(cos(0.876-2.09)+isin (0.876-2.09))# or

#Z=0.0969(cos 1.214-isin 1.214) =22/65-59/65i#

In trigonometric form: #0.969(cos 1.214-isin 1.214)# [Ans]