What is the trigonometric form of (5+10i)(3+3i) (5+10i)(3+3i)?

1 Answer
Dean R.
May 3, 2018

15 \sqrt{10} text{ cis} ( text{Arc}text{tan}(-3) ) 15√10 cis(Arctan(−3))

Explanation:

(5+10i)(3+3i) = 15(1+2i)(1+i) = 15(-1 + 3i)(5+10i)(3+3i)=15(1+2i)(1+i)=15(−1+3i)

= 15 \sqrt{1^2+3^2} (text{cis} ( arctan2(3 //, -1) ) }=15√12+32(cis(arctan2(3/,−1))}

I employ the funny notation because it take a two parameter, four quadrant arctangent in general. This one's in the fourth quadrant, so the principal value of the arctangent is sufficient:

= 15 \sqrt{10} text{ cis} ( text{Arc}text{tan}(-3) ) =15√10 cis(Arctan(−3))

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