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The value of the contour integral $$\oint\limits_{\left| {{\text{z}} - {\text{i}}} \right| = 2} {\frac{1}{{{{\text{z}}^2} + 4}}{\text{dz}}} $$    in positive sense is

The value of the contour integral $$\oint\limits_{\left| {{\text{z}} - {\text{i}}} \right| = 2} {\frac{1}{{{{\text{z}}^2} + 4}}{\text{dz}}} $$    in positive sense is Correct Answer $$\frac{\pi }{2}$$

Related Questions

Considered the following statements about the characteristics of contours - (1) Closed contour lines with higher values inside show a lake, (2) Contour is an imaginary line joining points of equal elevations, (3) Closely spaced contour lines indicate a steep slope. (4) Contour lines of equal height may not converge nor continue as one line but a single contour line may split into two lines Which of the statements given above are correct?
Consider the line integral $$I = \int_{\text{c}} {\left( {{{\text{x}}^2} + {\text{i}}{{\text{y}}^2}} \right){\text{dz,}}} $$    where z = x + iy. The line c is shown in the figure below
Complex Variable mcq question image
The value of $$I$$ is
The values of the integral $$\frac{1}{{2\pi {\text{j}}}}\oint\limits_{\text{c}} {\frac{{{{\text{e}}^{\text{z}}}}}{{{\text{z}} - 2}}{\text{dz}}} $$    along a closed contour c in anti-clockwise direction for
i. the point z0 = 2 inside the contour c, and
ii. the point z0 = 2 outside the contour c, respectively, are
If a + b + c + d = 4, then find the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?
If a + b + c + d = 4, then the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGjbGaaeOzaiaabckacaqG4bGaaeiOaiabgUcaRiaabckadaWc % aaWdaeaapeGaaGymaaWdaeaapeGaamiEaiabgkHiTiaaigdacaaIZa % aaaiabg2da9iaabckacaaIXaGaaGymaiaacYcacaqGGcGaaeiDaiaa % bIgacaqGLbGaaeOBaiaabckacaqGMbGaaeyAaiaab6gacaqGKbGaae % iOaiaabckacaqG0bGaaeiAaiaabwgacaqGGcGaaeODaiaabggacaqG % SbGaaeyDaiaabwgacaqGGcGaae4BaiaabAgacaqGGcWaaeWaa8aaba % WdbiaabIhacqGHsislcaaIXaGaaG4maaGaayjkaiaawMcaa8aadaah % aaWcbeqaa8qacaaI1aaaaOGaaeiOaiabgUcaRiaabckadaWcaaWdae % aapeGaaGymaaWdaeaapeWaaeWaa8aabaWdbiaadIhacqGHsislcaaI % XaGaaGymaaGaayjkaiaawMcaa8aadaahaaWcbeqaa8qacaaI1aaaaa % aakiaacckacaGGUaaaaa!70B8! {\rm{If\;x\;}} + {\rm{\;}}\frac{1}{{x - 13}} = {\rm{\;}}11,{\rm{\;then\;find\;\;the\;value\;of\;}}{\left( {{\rm{x}} - 13} \right)^5}{\rm{\;}} + {\rm{\;}}\frac{1}{{{{\left( {x - 11} \right)}^5}}}\;.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGjbGaaeOzaiaabckacaqG4bGaaeiOaiabgUcaRiaabckadaWc % aaWdaeaapeGaaGymaaWdaeaapeGaamiEaiabgkHiTiaaigdacaaIZa % aaaiabg2da9iaabckacaaIXaGaaGymaiaacYcacaqGGcGaaeiDaiaa % bIgacaqGLbGaaeOBaiaabckacaqGMbGaaeyAaiaab6gacaqGKbGaae % iOaiaabckacaqG0bGaaeiAaiaabwgacaqGGcGaaeODaiaabggacaqG % SbGaaeyDaiaabwgacaqGGcGaae4BaiaabAgacaqGGcWaaeWaa8aaba % WdbiaabIhacqGHsislcaaIXaGaaG4maaGaayjkaiaawMcaa8aadaah % aaWcbeqaa8qacaaI1aaaaOGaaeiOaiabgUcaRiaabckadaWcaaWdae % aapeGaaGymaaWdaeaapeWaaeWaa8aabaWdbiaadIhacqGHsislcaaI % XaGaaGymaaGaayjkaiaawMcaa8aadaahaaWcbeqaa8qacaaI1aaaaa % aakiaacckacaGGUaaaaa!70B8! {\rm{If\;x\;}} + {\rm{\;}}\frac{1}{{x - 13}} = {\rm{\;}}11,{\rm{\;then\;find\;\;the\;value\;of\;}}{\left( {{\rm{x}} - 13} \right)^5}{\rm{\;}} + {\rm{\;}}\frac{1}{{{{\left( {x - 11} \right)}^5}}}\;.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGjbGaaeOzaiaabckacaqG4bGaaeiOaiabgUcaRiaabckadaWc % aaWdaeaapeGaaGymaaWdaeaapeGaamiEaiabgkHiTiaaigdacaaIZa % aaaiabg2da9iaabckacaaIXaGaaGymaiaacYcacaqGGcGaaeiDaiaa % bIgacaqGLbGaaeOBaiaabckacaqGMbGaaeyAaiaab6gacaqGKbGaae % iOaiaabckacaqG0bGaaeiAaiaabwgacaqGGcGaaeODaiaabggacaqG % SbGaaeyDaiaabwgacaqGGcGaae4BaiaabAgacaqGGcWaaeWaa8aaba % WdbiaabIhacqGHsislcaaIXaGaaG4maaGaayjkaiaawMcaa8aadaah % aaWcbeqaa8qacaaI1aaaaOGaaeiOaiabgUcaRiaabckadaWcaaWdae % aapeGaaGymaaWdaeaapeWaaeWaa8aabaWdbiaadIhacqGHsislcaaI % XaGaaGymaaGaayjkaiaawMcaa8aadaahaaWcbeqaa8qacaaI1aaaaa % aakiaacckacaGGUaaaaa!70B8! {\rm{If\;x\;}} + {\rm{\;}}\frac{1}{{x - 13}} = {\rm{\;}}11,{\rm{\;then\;find\;\;the\;value\;of\;}}{\left( {{\rm{x}} - 13} \right)^5}{\rm{\;}} + {\rm{\;}}\frac{1}{{{{\left( {x - 11} \right)}^5}}}\;.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaqGjbGaaeOzaiaabckacaqG4bGaaeiOaiabgUcaRiaabckadaWc % aaWdaeaapeGaaGymaaWdaeaapeGaamiEaiabgkHiTiaaigdacaaIZa % aaaiabg2da9iaaigdacaaIXaGaaiilaiaabckacaqG0bGaaeiAaiaa % bwgacaqGUbGaaeiOaiaabAgacaqGPbGaaeOBaiaabsgacaqGGcGaae % iDaiaabIgacaqGLbGaaeiOaiaabAhacaqGHbGaaeiBaiaabwhacaqG % LbGaaeiOaiaab+gacaqGMbGaaeiOamaabmaapaqaa8qacaqG4bGaey % OeI0IaaGymaiaaiodaaiaawIcacaGLPaaapaWaaWbaaSqabeaapeGa % aGynaaaakiaabckacqGHRaWkcaqGGcWaaSaaa8aabaWdbiaaigdaa8 % aabaWdbmaabmaapaqaa8qacaWG4bGaeyOeI0IaaGymaiaaigdaaiaa % wIcacaGLPaaapaWaaWbaaSqabeaapeGaaGynaaaaaaGccaGGGcGaai % Olaaaa!6E72! {\rm{If\;x\;}} + {\rm{\;}}\frac{1}{{x - 13}} = 11,{\rm{\;then\;find\;the\;value\;of\;}}{\left( {{\rm{x}} - 13} \right)^5}{\rm{\;}} + {\rm{\;}}\frac{1}{{{{\left( {x - 11} \right)}^5}}}\;.\)If x + 1/(x - 13) = 11, then what will be the value of (x – 13)5 + 1/(x – 11)5?
If the semi-circular contour D of radius 2 is as shown in the figure, then the value of the integral $$\oint\limits_{\text{D}} {\frac{1}{{\left( {{{\text{s}}^2} - 1} \right)}}{\text{ds}}} $$   is
Complex Variable mcq question image
With respect to the numerical evaluation of the definite integral $${\text{K}} = \int_{\text{a}}^{\text{b}} {{{\text{x}}^2}{\text{dx,}}} $$   where a and b are given, which of the following statements is/are TRUE?
I. The value of K obtained using the trapezoidal rule is always greater than or equal to the exact value of the definite integral.
II. The value of K obtained using the Simpson's rule is always equal to the exact value of the definite integral.
What is correct sequence, of the following links from the Doctrine of Dependent Origination?
1. Trsna (thirst for sense enjoyment)
2. Vedana (sense experience)
3. Sparsa (sense-objet contact)
4. Upadana (clinging to sense enjoyment)
Select the correct answer:
The value of the contour integral in the complex plane $$\oint {\frac{{{{\text{z}}^3} - 2{\text{z}} + 3}}{{{\text{z}} - 2}}{\text{dz}}} $$    along the contour |z| = 3, taken counter-clockwise is