What is the trigonometric form of -8-i?

1 Answer
Feb 23, 2018

-(8+i)~~-sqrt58(cos(0.12)+isin(0.12))

Explanation:

-8-i=-(8+i)

For a given complex number, z=a+bi, z=r(costheta+isintheta)

r=sqrt(a^2+b^2)

theta=tan^-1(b/a)

Let's deal with 8+i

z=8+i=r(costheta+isintheta)

r=sqrt(8^2+1^2)=sqrt65

theta=tan^-1(1/8)~~0.12^c

-(8+i)~~-sqrt58(cos(0.12)+isin(0.12))