Q.
The value of Arg[i ln(a−iba+ib)] ; where a and b are real numbers
see full answer
High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET
🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya
a
0 or π
b
π2
c
Not defined
d
None of these
answer is A.
(Unlock A.I Detailed Solution for FREE)
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.
Watch 3-min video & get full concept clarity