http://google.com, pub-7771400403364887, DIRECT, f08c47fec0942fa0 obtain all the values of ln(√3-I) and find the principal value .

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obtain all the values of ln(√3-I) and find the principal value .

Solution .

Let z= √(3-i) then r= √{(3)²+(-1)²}=√4=2

And arg tan⁻¹(-1/√3)=360°-30°=330°=11π/6

Therefore lnz ={lnr +(0º+2πk)i, where k=0,±1,2,3...

Principal value is obtained when K = 0,lnz=ln2+(11π/6)i

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