How do you divide #( -4i+9) / (2i-12)# in trigonometric form?

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Answer 1 · 2016-02-06T15:33:41

#(58-3i)/74# More calculations are given below.

Multiply the denominator, with its conjugate. It is 2i+12 here. Now multiplication would work as follows:

#((-4i+9) ( 2i+12))/ ((-2i+12) (2i+12))#

= #(8 +108 -24i+18i)/(4+144)#

=#(116-6i)/148#

=#(58-3i)/74#

=#58/74 - i 3/74#

Now, let # r cos theta = 58/74# and #r sin theta= 3/74#

On squaring and adding #r^2=3373/5476# that means #r=sqrt3373/74# and on division it would be #tan theta=-3/58#, #theta= tan^-1 (-3/58)#

The required trignometric form would be #r(cos theta+isin theta)#, where r and #theta# would have values as worked out above.