How do you divide # (1+7i) / (9-5i) # in trigonometric form?

1 Answer
bp
Jun 4, 2016

The result of division is #1/53( -13+34i)#
Its trignometrical form #(sqrt1325)/53 (-13/sqrt1325 +34/sqrt1325 i)#

Explanation:

First of all multiply the numerator and the denominator by the conjugate of denominator.

#((1+7i)(9+5i))/((9-5i)(9+5i))= (-26+68i)/106#

=#1/53( -13+34i)#

Now Modulus would be #sqrt (13^2 +34^2)=sqrt1325#

The number can now be written as #(sqrt1325)/53 (-13/sqrt1325 +34/sqrt1325 i)#

The number now is in trignometrical form its Modulus is #sqrt1325 /53# and arg is #tan^-1 (34/(-13))#

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