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

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Answer 1 · 2016-06-04T03:23:10

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

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 $tan^-1 (34/(-13))$

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