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

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Answer 1 · 2017-07-17T17:18:11

#(-4+9i)/(1-5i)=1/26sqrt2522(cos167-isin167)#

#(-4+9i)/(1-5i)=((-4+9i)(1+5i))/((1-5i)(1+5i))=(-4-11i-45)/(26)=(-11i-49)/26=-49/26-11/26i#

Trigonometric form of a complex number:

#a+bi=r(cosvartheta+isinvartheta)#, where #r=sqrt(a^2+b^2)# and #vartheta=arctan(b/a)#

#sqrt(49/26^2+11/26^2)=1/26sqrt2522#

#arctan((-11/26)/(-49/26))approx-167^@#

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

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