complex..

wat is arg[(3^1/2+i)/1-i]?? pls give me ans 2 check..

7 Answers

(3^1/2+i)/(1-i)

(√3+i)(1+i)/2

(√3-1)/2+i(√3+1)/2

now the arguement is?

thats right so could u pls give me the final ans....

tan^-1(2+√3)

or equivalently
tan^-1[(âˆš3+1)/(âˆš3-1)]

=tan^-1[(1+1/âˆš3)/(1-1/âˆš3)]
=tan^-1[tan(Ï€/4+Ï€/6)]
=Ï€/4+Ï€/6
=5Ï€/12

OOppsss.. this is much better!

Gr8 work Prashant

Yes this is another good way to solve this one :)

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