Q.

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.
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