The value of Arg[i ln(a−iba+ib)] ; where a and b are real numbers
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a
0 or π
b
π2
c
Not defined
d
None of these
answer is A.
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Detailed Solution
ln(a−iba+ib)=ln|a−iba+ib|+i[2nπ+arg(a−iba+ib)] =i[2nπ+arg(a−iba+ib)] ∵|a−iba+ib|=1 ∴Arg[iln(a−iba+ib)]=Argi[−2nπ−arg(a−iba+ib)]=0 or πAs 2nπ+arg(a−iba+ib) is a real number.