What is the trigonometric form of #-8-i#?

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Answer 1 · 2018-02-23T10:20:17

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

#-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))#