How do you divide (1+7i) / (9-5i) 1+7i9−5i in trigonometric form?

1 Answer
bp
Jun 4, 2016

The result of division is 1/53( -13+34i)153(−13+34i)
Its trignometrical form (sqrt1325)/53 (-13/sqrt1325 +34/sqrt1325 i)√132553(−13√1325+34√1325i)

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+7i)(9+5i)(9−5i)(9+5i)=−26+68i106

=1/53( -13+34i)153(−13+34i)

Now Modulus would be sqrt (13^2 +34^2)=sqrt1325√132+342=√1325

The number can now be written as (sqrt1325)/53 (-13/sqrt1325 +34/sqrt1325 i)√132553(−13√1325+34√1325i)

The number now is in trignometrical form its Modulus is sqrt1325 /53√132553 and arg is tan^-1 (34/(-13))tan−1(34−13)

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