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

Question

1514 views around the world

You can reuse this answer
Creative Commons License

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 $1/26sqrt2522(cos167-isin167)$

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