How do you divide 2i−45i−6 in trigonometric form?

1 Answer
Shiva Prakash M V
Feb 18, 2018

2i−45i−6=√561cis(tan−1(−1629))

Explanation:

Letz1=2i−4,where,
r1=√22+(−1)2=√5
θ1=tan−1(−42)

Letz2=5i−6,where,
r1=√52+(−6)2=√61
θ1=tan−1(−65)

Now,

z1=√5(cis(tan−1(−42)))

z2=√61(cis(tan−1(−65)))

2i−45i−6=z1z2=√5(cis(tan−1(−42)))√61(cis(tan−1(−65)))
=√5√61cis(tan−1(−42))cis(tan−1(−65))

√5√61=√561
By De-Moivre's Theorem,

cis(tan−1(−42))cis(tan−1(−65))=cis(tan−1(−42)−tan−1(−65))

tan−1(−42)−tan−1(−65)=tan−1⎛⎜⎝−42−−651+(−42)(−65)⎞⎟⎠

−42−65=−2−65=−10−65=−165

1+(−42)(−65)=1+245=5+245=295

Hence,
tan−1⎛⎜⎝−42−−651+(−42)(−65)⎞⎟⎠=tan−1(−165295)=tan−1(−1629)

Thus,

2i−45i−6=√561cis(tan−1(−1629))

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