How do you divide #(9i-5) / (-2i+6)# in trigonometric form?

1 Answer
Dean R.
Apr 23, 2018

# frac{-5 +9i}{6-2i} = {-12+11i}/10# but I couldn't finish in trigonometric form.

Explanation:

These are nice complex numbers in rectangular form. It's a big waste of time to convert them to polar coordinates to divide them. Let's try it both ways:

# frac{-5 +9i}{6-2i} cdot {6+2i}/{6+2i} = {-48 +44i }/{40} = {-12+11i}/10 #

That was easy. Let's contrast.

In polar coordinates we have

#-5 + 9i = \sqrt{5^2+9^2} \ e^{i text{ atan2}(9,-5) } #

I write #text{atan2}(y,x)# as the correct two parameter, four quadrant inverse tangent.

# 6-2i = \sqrt{6^2+2^2} \ e^{i text{ atan2}(-2, 6) } #

# frac{-5 +9i}{6-2i} = frac{ \sqrt{106 } e^{ i text{ atan2}(9,-5)} }{ \sqrt{40} \ e^{i text{ atan2}(-2, 6) }}#

# frac{-5 +9i}{6-2i} = \sqrt{106/40} e^{ i ( text{ atan2}(9,-5) - text{atan2}(-2, 6)) } #

We can actually make progress with the tangent difference angle formula, but I'm not up for that. I suppose we could get the calculator out, but why turn a nice exact problem into an approximation?

Uncle.

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