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.

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