How do you divide # (6+9i)/(1+i) # in trigonometric form?

1 Answer
t0hierry
Apr 5, 2016

Sorry for the delay. Below is my answer.

Explanation:

First step, get rid of complex denominator by multiplying top and bottom by its complex conjugate
#r = (6 + 9i)(1 - i)/[(1 + i)(1-i)] = (6 + 9i - 6i + 9)/2 = 15/2 + 3i/2#

Next you factor out the modulus or norm of the complex number. and write

#r = sqrt(225 + 9)/2(cos alpha + i sin alpha)#

where #alpha# is such that #tan alpha = 1/5#

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