How do you divide (-4+9i)/(1-5i) −4+9i1−5i in trigonometric form?

1 Answer
Monzur R.
Jul 17, 2017

(-4+9i)/(1-5i)=1/26sqrt2522(cos167-isin167)−4+9i1−5i=126√2522(cos167−isin167)

Explanation:

(-4+9i)/(1-5i)=((-4+9i)(1+5i))/((1-5i)(1+5i))=(-4-11i-45)/(26)=(-11i-49)/26=-49/26-11/26i−4+9i1−5i=(−4+9i)(1+5i)(1−5i)(1+5i)=−4−11i−4526=−11i−4926=−4926−1126i

Trigonometric form of a complex number:

a+bi=r(cosvartheta+isinvartheta)a+bi=r(cosϑ+isinϑ), where r=sqrt(a^2+b^2)r=√a2+b2 and vartheta=arctan(b/a)ϑ=arctan(ba)

sqrt(49/26^2+11/26^2)=1/26sqrt2522√49262+11262=126√2522

arctan((-11/26)/(-49/26))approx-167^@arctan(−1126−4926)≈−167∘

therefore-49/26-11/26i=1/26sqrt2522(cos(-167)+isin(-167))

which can be rewritten as 1/26sqrt2522(cos167-isin167)

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