Question

8. Residue of $f(z) = \frac{\log(1+z)}{(1+z^2)^2}$ at $z = i$ is given by (A) $\frac{1}{8} - \frac{\pi}{16} + i \left[ -\frac{1}{8} + \frac{1}{4} \log\sqrt{2} \right]$ (B) $-\frac{1}{8} + \frac{\pi}{16} + i \left[ \frac{1}{8} - \frac{1}{4} \log\sqrt{2} \right]$ (C) $-\frac{1}{8} + \frac{\pi}{16} + i \left[ -\frac{1}{8} + \frac{1}{4} \log\sqrt{2} \right]$ (D) $\frac{1}{8} - \frac{\pi}{16} + i \left[ \frac{1}{8} - \frac{1}{4} \log\sqrt{2} \right]$

          8. Residue of $f(z) = \frac{\log(1+z)}{(1+z^2)^2}$ at $z = i$ is given by
(A) $\frac{1}{8} - \frac{\pi}{16} + i \left[ -\frac{1}{8} + \frac{1}{4} \log\sqrt{2} \right]$
(B) $-\frac{1}{8} + \frac{\pi}{16} + i \left[ \frac{1}{8} - \frac{1}{4} \log\sqrt{2} \right]$
(C) $-\frac{1}{8} + \frac{\pi}{16} + i \left[ -\frac{1}{8} + \frac{1}{4} \log\sqrt{2} \right]$
(D) $\frac{1}{8} - \frac{\pi}{16} + i \left[ \frac{1}{8} - \frac{1}{4} \log\sqrt{2} \right]$

8. Residue of f(z) = (log(1+z))/((1+z^2)^2) at z = i is given by
(A) (1)/(8) - (π)/(16) + i [ -(1)/(8) + (1)/(4)log√(2)]
(B) -(1)/(8) + (π)/(16) + i [ (1)/(8) - (1)/(4)log√(2)]
(C) -(1)/(8) + (π)/(16) + i [ -(1)/(8) + (1)/(4)log√(2)]
(D) (1)/(8) - (π)/(16) + i [ (1)/(8) - (1)/(4)log√(2)]

Added by Michael B.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Madhur L

06/16/2023

Step 1

To do this, we can use the formula for the residue at a simple pole: Res(f(z), z=i) = lim(z→i) (z-i) * f(z) Show more…

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Residue of f(z)=(log(1+z))/((1+z^(2))^(2)) at z=i is given by (A) (1)/(8)-(pi )/(16)+i[-(1)/(8)+(1)/(4)logsqrt(2)] (B) -(1)/(8)+(pi )/(16)+i[(1)/(8)-(1)/(4)logsqrt(2)] (C) -(1)/(8)+(pi )/(16)+i[-(1)/(8)+(1)/(4)logsqrt(2)] (D) (1)/(8)-(pi )/(16)+i[(1)/(8)-(1)/(4)logsqrt(2)] log(1+z) at z=iis givenby 8. Residue of f(z) 1 T (A) 8 16 8 TT 16 1 8 (B) 8 4 1 TT 18 16 1 18 (C) 1 T (D) 8 16 1 8 4