UPSC Exam > UPSC Notes > Mathematics Optional Notes for UPSC > Analytic Functions
Analytic Functions | Mathematics Optional Notes for UPSC PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1 Edurev123 Complex Analysis 1. Analytic Functions 1.1 Let ?? (?? )= ?? ?? +?? ?? ?? +?.+?? ?? -?? ?? ?? -?? ?? ?? +?? ?? ?? +?.+?? ?? ?? ?? ,?? ?? ??? . Assume that the zeroes of the denominator are simple. Show that the sum of the residues of ?? (?? ) at its poles is equal to ?? ?? -?? ?? ?? . (2009 : 12 Marks) Solution: Proof: As the zeroes of denominator are simple, so ?? 0 +?? 1 ?? +?..+?? ?? ?? ?? can be written as ?? ?? (?? -?? 0 )(?? -?? 1 )(?? -?? 2 )…(?? -?? ?? -1 ) or As maximum degree of ' ?? ' in numerator is ?? -1 and degree of denominator is ?? (i.e., degree of denominator is greater than degree of numerator). So, ?? (?? ) can be written as ?? (?? )= ?? 0 ?? -?? 0 + ?? 1 ?? -?? 1 + ?? 2 ?? -?? 2 +?… ?? ?? -1 ?? -?? ?? -1 (2) = ?? 0 (?? -?? 1 )(?? -2)·….(?? -?? ?? -1 )+?? 1 (?? -?? 0 )(?? -?? 2 )-(?? -?? ?? -1 )+??? ?? (?? -?? 0 )(?? -?? ?? -2 ) (?? -?? 0 )(?? -?? 1 )…..(?? -?? ?? -1 ) (3) Now, we know that residues of ?? (?? ) are ?? 0 ,?? 1 ,?? 2 ,….,?? ?? -1 . According to problem, we need to calculate ?? 0 +?? 1 +?? 2 …..?? ?? -1 . From (3), the coefficient of ?? ?? -1 in numerator will be calculated as : ?? 0 +?? 1 +?? 2 …..?? ?? -1 because on observing ?? 0 (?? -?? 1 )(?? -?? 2 )-(?? -?? ?? -1 ) +?? 1 (?? -?? 0 )(?? -?? 2 )-(?? -?? ?? -1 ) +?? 2 (?? -?? 0 )(?? -?? 1 )(?? -?? 3 )-(?? -?? ?? -1 ) +?? +?? ?? -1 (?? -?? 0 )(?? -?? 1 )….(?? -?? ?? -2 ) Page 2 Edurev123 Complex Analysis 1. Analytic Functions 1.1 Let ?? (?? )= ?? ?? +?? ?? ?? +?.+?? ?? -?? ?? ?? -?? ?? ?? +?? ?? ?? +?.+?? ?? ?? ?? ,?? ?? ??? . Assume that the zeroes of the denominator are simple. Show that the sum of the residues of ?? (?? ) at its poles is equal to ?? ?? -?? ?? ?? . (2009 : 12 Marks) Solution: Proof: As the zeroes of denominator are simple, so ?? 0 +?? 1 ?? +?..+?? ?? ?? ?? can be written as ?? ?? (?? -?? 0 )(?? -?? 1 )(?? -?? 2 )…(?? -?? ?? -1 ) or As maximum degree of ' ?? ' in numerator is ?? -1 and degree of denominator is ?? (i.e., degree of denominator is greater than degree of numerator). So, ?? (?? ) can be written as ?? (?? )= ?? 0 ?? -?? 0 + ?? 1 ?? -?? 1 + ?? 2 ?? -?? 2 +?… ?? ?? -1 ?? -?? ?? -1 (2) = ?? 0 (?? -?? 1 )(?? -2)·….(?? -?? ?? -1 )+?? 1 (?? -?? 0 )(?? -?? 2 )-(?? -?? ?? -1 )+??? ?? (?? -?? 0 )(?? -?? ?? -2 ) (?? -?? 0 )(?? -?? 1 )…..(?? -?? ?? -1 ) (3) Now, we know that residues of ?? (?? ) are ?? 0 ,?? 1 ,?? 2 ,….,?? ?? -1 . According to problem, we need to calculate ?? 0 +?? 1 +?? 2 …..?? ?? -1 . From (3), the coefficient of ?? ?? -1 in numerator will be calculated as : ?? 0 +?? 1 +?? 2 …..?? ?? -1 because on observing ?? 0 (?? -?? 1 )(?? -?? 2 )-(?? -?? ?? -1 ) +?? 1 (?? -?? 0 )(?? -?? 2 )-(?? -?? ?? -1 ) +?? 2 (?? -?? 0 )(?? -?? 1 )(?? -?? 3 )-(?? -?? ?? -1 ) +?? +?? ?? -1 (?? -?? 0 )(?? -?? 1 )….(?? -?? ?? -2 ) In each of terms respectively coefficient of ?? ?? -1 will be ?? 0 ,?? 1 ,?? 2 …..?? ?? -1 . So, the overall coefficient of ?? ?? -1 in numerator will be ?? 0 +?? 1 +?? 2 …..?? ?? -1 . Now, comparing this coefficient of 2 ?? -1 with (1) i.e., ?? 0 ?? ?? + ?? 1 ?? ?? ?? + ?? 2 ?? ?? ?? 2 ….· ?? ?? -1 ?? ?? ?? ?? -1 (?? -?? 0 )(?? -?? 1 )….(?? -?? ?? -1 ) (4) In this fraction, coefficient of ?? ?? -1 will be ?? ?? -1 ?? ?? . (as denominator (?? -?? 0 )(?? -?? 1 )…..(?? -?? ?? -1 ) is common (3) and (4)) ??????????????????????????????????? 0 +?? 1 +?? 2 …?? ?? -1 = ?? ?? -1 ?? ?? or Sum of residues = ?? ?? -1 ?? ?? . Thus Proved. 1.2 Show that ?? (?? ,?? )=?? ?? -?? ?? +?? ?? ?? ?? is a harmonic function. Find a harmonic conjugate of ?? (?? ,?? ) . Hence, find the analytic function ?? for which ?? (?? ,?? ) is the real part. (2010 : 12 Marks) Solution: Given ?? (?? ,?? )?=2?? -?? 3 +3?? ?? 2 ??? ??? ?=2-3?? 2 +3?? 2 , ? 2 ?? ??? 2 =-6?? ??? ??? ?=6???? , ? 2 ?? ??? 2 =6?? ??????????????????????????????????????????? ? 2 ?? ??? 2 + ? 2 ?? ??? 2 =-6?? +6?? =0 ??? (?? ,?? ) is a harmonic function. Let ?? be its harmonic conjugate. Page 3 Edurev123 Complex Analysis 1. Analytic Functions 1.1 Let ?? (?? )= ?? ?? +?? ?? ?? +?.+?? ?? -?? ?? ?? -?? ?? ?? +?? ?? ?? +?.+?? ?? ?? ?? ,?? ?? ??? . Assume that the zeroes of the denominator are simple. Show that the sum of the residues of ?? (?? ) at its poles is equal to ?? ?? -?? ?? ?? . (2009 : 12 Marks) Solution: Proof: As the zeroes of denominator are simple, so ?? 0 +?? 1 ?? +?..+?? ?? ?? ?? can be written as ?? ?? (?? -?? 0 )(?? -?? 1 )(?? -?? 2 )…(?? -?? ?? -1 ) or As maximum degree of ' ?? ' in numerator is ?? -1 and degree of denominator is ?? (i.e., degree of denominator is greater than degree of numerator). So, ?? (?? ) can be written as ?? (?? )= ?? 0 ?? -?? 0 + ?? 1 ?? -?? 1 + ?? 2 ?? -?? 2 +?… ?? ?? -1 ?? -?? ?? -1 (2) = ?? 0 (?? -?? 1 )(?? -2)·….(?? -?? ?? -1 )+?? 1 (?? -?? 0 )(?? -?? 2 )-(?? -?? ?? -1 )+??? ?? (?? -?? 0 )(?? -?? ?? -2 ) (?? -?? 0 )(?? -?? 1 )…..(?? -?? ?? -1 ) (3) Now, we know that residues of ?? (?? ) are ?? 0 ,?? 1 ,?? 2 ,….,?? ?? -1 . According to problem, we need to calculate ?? 0 +?? 1 +?? 2 …..?? ?? -1 . From (3), the coefficient of ?? ?? -1 in numerator will be calculated as : ?? 0 +?? 1 +?? 2 …..?? ?? -1 because on observing ?? 0 (?? -?? 1 )(?? -?? 2 )-(?? -?? ?? -1 ) +?? 1 (?? -?? 0 )(?? -?? 2 )-(?? -?? ?? -1 ) +?? 2 (?? -?? 0 )(?? -?? 1 )(?? -?? 3 )-(?? -?? ?? -1 ) +?? +?? ?? -1 (?? -?? 0 )(?? -?? 1 )….(?? -?? ?? -2 ) In each of terms respectively coefficient of ?? ?? -1 will be ?? 0 ,?? 1 ,?? 2 …..?? ?? -1 . So, the overall coefficient of ?? ?? -1 in numerator will be ?? 0 +?? 1 +?? 2 …..?? ?? -1 . Now, comparing this coefficient of 2 ?? -1 with (1) i.e., ?? 0 ?? ?? + ?? 1 ?? ?? ?? + ?? 2 ?? ?? ?? 2 ….· ?? ?? -1 ?? ?? ?? ?? -1 (?? -?? 0 )(?? -?? 1 )….(?? -?? ?? -1 ) (4) In this fraction, coefficient of ?? ?? -1 will be ?? ?? -1 ?? ?? . (as denominator (?? -?? 0 )(?? -?? 1 )…..(?? -?? ?? -1 ) is common (3) and (4)) ??????????????????????????????????? 0 +?? 1 +?? 2 …?? ?? -1 = ?? ?? -1 ?? ?? or Sum of residues = ?? ?? -1 ?? ?? . Thus Proved. 1.2 Show that ?? (?? ,?? )=?? ?? -?? ?? +?? ?? ?? ?? is a harmonic function. Find a harmonic conjugate of ?? (?? ,?? ) . Hence, find the analytic function ?? for which ?? (?? ,?? ) is the real part. (2010 : 12 Marks) Solution: Given ?? (?? ,?? )?=2?? -?? 3 +3?? ?? 2 ??? ??? ?=2-3?? 2 +3?? 2 , ? 2 ?? ??? 2 =-6?? ??? ??? ?=6???? , ? 2 ?? ??? 2 =6?? ??????????????????????????????????????????? ? 2 ?? ??? 2 + ? 2 ?? ??? 2 =-6?? +6?? =0 ??? (?? ,?? ) is a harmonic function. Let ?? be its harmonic conjugate. ? ??? ??? = ??? ??? =2-3?? 2 +3?? 2 ? ?? =2?? -3?? 2 ?? +?? 3 +?? (?? ) Also, ??? ??? =- ??? ??? =6???? ? ??? ??? =-6???? ? ?? =-3?? 2 ?? +?? (?? )…(2)(?? (?? ) and ?? (?? ) are some arbitrary functions) (1) Comparing (1) and (2), we get Now, let ?? ?=2?? +?? 3 -3?? 2 ?? +?? , where ?? is a constant. ?? ?=?? (?? ,?? )+???? (?? ,?? ) ?? ?=(2?? -?? 3 +3?? ?? 2 )+??(2?? +?? 3 -3?? 2 ?? +?? ) ?? (?? )?=??(?? 1 (?? ,0)-???? 2 (?? ,0))???? ?=??(2-3?? 2 )???? =2?? -?? 3 +?? , where ?? is a constant ???? ?=(2?? -?? 3 +3) ?? (?? )?=??(?? 1 (?? ,0)-?? ?=??(2-3?? 2 )?? ???? (?? )?=2?? -?? 3 +?? 1.3 (i) Evaluate the line integral ? ?? ??? (?? )???? where ?? (?? )=?? ?? ,?? is the boundary of the triangle with vertices ?? (?? ,?? ),?? (?? ,?? ),?? (?? ,?? ) in that order. (ii) Find the image of the finite vertical strip ?? : ?? =?? to ?? =?? ,-?? =?? =?? of ?? - plane under exponential function. (2010 : 15 Marks) Solution: (i) Given: ?? (?? )=?? 2 Now, ?? (?? )=?? 2 is analytic in whole given region. Line integral of cled boundary curve of analytic function is always zero. ?? ?? ??? (?? )???? in the given region is 0 . (ii) Given : ?? varies from 5 to 9 and ?? varies from -?? to ?? . Now, ??? =?? +???? Page 4 Edurev123 Complex Analysis 1. Analytic Functions 1.1 Let ?? (?? )= ?? ?? +?? ?? ?? +?.+?? ?? -?? ?? ?? -?? ?? ?? +?? ?? ?? +?.+?? ?? ?? ?? ,?? ?? ??? . Assume that the zeroes of the denominator are simple. Show that the sum of the residues of ?? (?? ) at its poles is equal to ?? ?? -?? ?? ?? . (2009 : 12 Marks) Solution: Proof: As the zeroes of denominator are simple, so ?? 0 +?? 1 ?? +?..+?? ?? ?? ?? can be written as ?? ?? (?? -?? 0 )(?? -?? 1 )(?? -?? 2 )…(?? -?? ?? -1 ) or As maximum degree of ' ?? ' in numerator is ?? -1 and degree of denominator is ?? (i.e., degree of denominator is greater than degree of numerator). So, ?? (?? ) can be written as ?? (?? )= ?? 0 ?? -?? 0 + ?? 1 ?? -?? 1 + ?? 2 ?? -?? 2 +?… ?? ?? -1 ?? -?? ?? -1 (2) = ?? 0 (?? -?? 1 )(?? -2)·….(?? -?? ?? -1 )+?? 1 (?? -?? 0 )(?? -?? 2 )-(?? -?? ?? -1 )+??? ?? (?? -?? 0 )(?? -?? ?? -2 ) (?? -?? 0 )(?? -?? 1 )…..(?? -?? ?? -1 ) (3) Now, we know that residues of ?? (?? ) are ?? 0 ,?? 1 ,?? 2 ,….,?? ?? -1 . According to problem, we need to calculate ?? 0 +?? 1 +?? 2 …..?? ?? -1 . From (3), the coefficient of ?? ?? -1 in numerator will be calculated as : ?? 0 +?? 1 +?? 2 …..?? ?? -1 because on observing ?? 0 (?? -?? 1 )(?? -?? 2 )-(?? -?? ?? -1 ) +?? 1 (?? -?? 0 )(?? -?? 2 )-(?? -?? ?? -1 ) +?? 2 (?? -?? 0 )(?? -?? 1 )(?? -?? 3 )-(?? -?? ?? -1 ) +?? +?? ?? -1 (?? -?? 0 )(?? -?? 1 )….(?? -?? ?? -2 ) In each of terms respectively coefficient of ?? ?? -1 will be ?? 0 ,?? 1 ,?? 2 …..?? ?? -1 . So, the overall coefficient of ?? ?? -1 in numerator will be ?? 0 +?? 1 +?? 2 …..?? ?? -1 . Now, comparing this coefficient of 2 ?? -1 with (1) i.e., ?? 0 ?? ?? + ?? 1 ?? ?? ?? + ?? 2 ?? ?? ?? 2 ….· ?? ?? -1 ?? ?? ?? ?? -1 (?? -?? 0 )(?? -?? 1 )….(?? -?? ?? -1 ) (4) In this fraction, coefficient of ?? ?? -1 will be ?? ?? -1 ?? ?? . (as denominator (?? -?? 0 )(?? -?? 1 )…..(?? -?? ?? -1 ) is common (3) and (4)) ??????????????????????????????????? 0 +?? 1 +?? 2 …?? ?? -1 = ?? ?? -1 ?? ?? or Sum of residues = ?? ?? -1 ?? ?? . Thus Proved. 1.2 Show that ?? (?? ,?? )=?? ?? -?? ?? +?? ?? ?? ?? is a harmonic function. Find a harmonic conjugate of ?? (?? ,?? ) . Hence, find the analytic function ?? for which ?? (?? ,?? ) is the real part. (2010 : 12 Marks) Solution: Given ?? (?? ,?? )?=2?? -?? 3 +3?? ?? 2 ??? ??? ?=2-3?? 2 +3?? 2 , ? 2 ?? ??? 2 =-6?? ??? ??? ?=6???? , ? 2 ?? ??? 2 =6?? ??????????????????????????????????????????? ? 2 ?? ??? 2 + ? 2 ?? ??? 2 =-6?? +6?? =0 ??? (?? ,?? ) is a harmonic function. Let ?? be its harmonic conjugate. ? ??? ??? = ??? ??? =2-3?? 2 +3?? 2 ? ?? =2?? -3?? 2 ?? +?? 3 +?? (?? ) Also, ??? ??? =- ??? ??? =6???? ? ??? ??? =-6???? ? ?? =-3?? 2 ?? +?? (?? )…(2)(?? (?? ) and ?? (?? ) are some arbitrary functions) (1) Comparing (1) and (2), we get Now, let ?? ?=2?? +?? 3 -3?? 2 ?? +?? , where ?? is a constant. ?? ?=?? (?? ,?? )+???? (?? ,?? ) ?? ?=(2?? -?? 3 +3?? ?? 2 )+??(2?? +?? 3 -3?? 2 ?? +?? ) ?? (?? )?=??(?? 1 (?? ,0)-???? 2 (?? ,0))???? ?=??(2-3?? 2 )???? =2?? -?? 3 +?? , where ?? is a constant ???? ?=(2?? -?? 3 +3) ?? (?? )?=??(?? 1 (?? ,0)-?? ?=??(2-3?? 2 )?? ???? (?? )?=2?? -?? 3 +?? 1.3 (i) Evaluate the line integral ? ?? ??? (?? )???? where ?? (?? )=?? ?? ,?? is the boundary of the triangle with vertices ?? (?? ,?? ),?? (?? ,?? ),?? (?? ,?? ) in that order. (ii) Find the image of the finite vertical strip ?? : ?? =?? to ?? =?? ,-?? =?? =?? of ?? - plane under exponential function. (2010 : 15 Marks) Solution: (i) Given: ?? (?? )=?? 2 Now, ?? (?? )=?? 2 is analytic in whole given region. Line integral of cled boundary curve of analytic function is always zero. ?? ?? ??? (?? )???? in the given region is 0 . (ii) Given : ?? varies from 5 to 9 and ?? varies from -?? to ?? . Now, ??? =?? +???? Image under exponential function will be ?? ?? =?? ?? +???? =?? ?? ·?? ???? As ?? varies from 5 to 9,??? ?? varies from ?? 5 to ?? 9 . Also, ?? ??4 denotes angle variation. So, image is given by region in figure. So, the region is bounded by two circles of radii ?? 5 and ?? 9 . 1.4 If ?? (?? )=?? +?? is an analytic function of ?? =?? +?? and ?? -?? = ?? ?? -?????? ??? +?????? ??? ???????? ??? -?????? ??? , find ?? (?? ) subject to the condition, ?? ( ?? ?? )= ?? -?? ?? . (2011 : 12 Marks) Solution: Page 5 Edurev123 Complex Analysis 1. Analytic Functions 1.1 Let ?? (?? )= ?? ?? +?? ?? ?? +?.+?? ?? -?? ?? ?? -?? ?? ?? +?? ?? ?? +?.+?? ?? ?? ?? ,?? ?? ??? . Assume that the zeroes of the denominator are simple. Show that the sum of the residues of ?? (?? ) at its poles is equal to ?? ?? -?? ?? ?? . (2009 : 12 Marks) Solution: Proof: As the zeroes of denominator are simple, so ?? 0 +?? 1 ?? +?..+?? ?? ?? ?? can be written as ?? ?? (?? -?? 0 )(?? -?? 1 )(?? -?? 2 )…(?? -?? ?? -1 ) or As maximum degree of ' ?? ' in numerator is ?? -1 and degree of denominator is ?? (i.e., degree of denominator is greater than degree of numerator). So, ?? (?? ) can be written as ?? (?? )= ?? 0 ?? -?? 0 + ?? 1 ?? -?? 1 + ?? 2 ?? -?? 2 +?… ?? ?? -1 ?? -?? ?? -1 (2) = ?? 0 (?? -?? 1 )(?? -2)·….(?? -?? ?? -1 )+?? 1 (?? -?? 0 )(?? -?? 2 )-(?? -?? ?? -1 )+??? ?? (?? -?? 0 )(?? -?? ?? -2 ) (?? -?? 0 )(?? -?? 1 )…..(?? -?? ?? -1 ) (3) Now, we know that residues of ?? (?? ) are ?? 0 ,?? 1 ,?? 2 ,….,?? ?? -1 . According to problem, we need to calculate ?? 0 +?? 1 +?? 2 …..?? ?? -1 . From (3), the coefficient of ?? ?? -1 in numerator will be calculated as : ?? 0 +?? 1 +?? 2 …..?? ?? -1 because on observing ?? 0 (?? -?? 1 )(?? -?? 2 )-(?? -?? ?? -1 ) +?? 1 (?? -?? 0 )(?? -?? 2 )-(?? -?? ?? -1 ) +?? 2 (?? -?? 0 )(?? -?? 1 )(?? -?? 3 )-(?? -?? ?? -1 ) +?? +?? ?? -1 (?? -?? 0 )(?? -?? 1 )….(?? -?? ?? -2 ) In each of terms respectively coefficient of ?? ?? -1 will be ?? 0 ,?? 1 ,?? 2 …..?? ?? -1 . So, the overall coefficient of ?? ?? -1 in numerator will be ?? 0 +?? 1 +?? 2 …..?? ?? -1 . Now, comparing this coefficient of 2 ?? -1 with (1) i.e., ?? 0 ?? ?? + ?? 1 ?? ?? ?? + ?? 2 ?? ?? ?? 2 ….· ?? ?? -1 ?? ?? ?? ?? -1 (?? -?? 0 )(?? -?? 1 )….(?? -?? ?? -1 ) (4) In this fraction, coefficient of ?? ?? -1 will be ?? ?? -1 ?? ?? . (as denominator (?? -?? 0 )(?? -?? 1 )…..(?? -?? ?? -1 ) is common (3) and (4)) ??????????????????????????????????? 0 +?? 1 +?? 2 …?? ?? -1 = ?? ?? -1 ?? ?? or Sum of residues = ?? ?? -1 ?? ?? . Thus Proved. 1.2 Show that ?? (?? ,?? )=?? ?? -?? ?? +?? ?? ?? ?? is a harmonic function. Find a harmonic conjugate of ?? (?? ,?? ) . Hence, find the analytic function ?? for which ?? (?? ,?? ) is the real part. (2010 : 12 Marks) Solution: Given ?? (?? ,?? )?=2?? -?? 3 +3?? ?? 2 ??? ??? ?=2-3?? 2 +3?? 2 , ? 2 ?? ??? 2 =-6?? ??? ??? ?=6???? , ? 2 ?? ??? 2 =6?? ??????????????????????????????????????????? ? 2 ?? ??? 2 + ? 2 ?? ??? 2 =-6?? +6?? =0 ??? (?? ,?? ) is a harmonic function. Let ?? be its harmonic conjugate. ? ??? ??? = ??? ??? =2-3?? 2 +3?? 2 ? ?? =2?? -3?? 2 ?? +?? 3 +?? (?? ) Also, ??? ??? =- ??? ??? =6???? ? ??? ??? =-6???? ? ?? =-3?? 2 ?? +?? (?? )…(2)(?? (?? ) and ?? (?? ) are some arbitrary functions) (1) Comparing (1) and (2), we get Now, let ?? ?=2?? +?? 3 -3?? 2 ?? +?? , where ?? is a constant. ?? ?=?? (?? ,?? )+???? (?? ,?? ) ?? ?=(2?? -?? 3 +3?? ?? 2 )+??(2?? +?? 3 -3?? 2 ?? +?? ) ?? (?? )?=??(?? 1 (?? ,0)-???? 2 (?? ,0))???? ?=??(2-3?? 2 )???? =2?? -?? 3 +?? , where ?? is a constant ???? ?=(2?? -?? 3 +3) ?? (?? )?=??(?? 1 (?? ,0)-?? ?=??(2-3?? 2 )?? ???? (?? )?=2?? -?? 3 +?? 1.3 (i) Evaluate the line integral ? ?? ??? (?? )???? where ?? (?? )=?? ?? ,?? is the boundary of the triangle with vertices ?? (?? ,?? ),?? (?? ,?? ),?? (?? ,?? ) in that order. (ii) Find the image of the finite vertical strip ?? : ?? =?? to ?? =?? ,-?? =?? =?? of ?? - plane under exponential function. (2010 : 15 Marks) Solution: (i) Given: ?? (?? )=?? 2 Now, ?? (?? )=?? 2 is analytic in whole given region. Line integral of cled boundary curve of analytic function is always zero. ?? ?? ??? (?? )???? in the given region is 0 . (ii) Given : ?? varies from 5 to 9 and ?? varies from -?? to ?? . Now, ??? =?? +???? Image under exponential function will be ?? ?? =?? ?? +???? =?? ?? ·?? ???? As ?? varies from 5 to 9,??? ?? varies from ?? 5 to ?? 9 . Also, ?? ??4 denotes angle variation. So, image is given by region in figure. So, the region is bounded by two circles of radii ?? 5 and ?? 9 . 1.4 If ?? (?? )=?? +?? is an analytic function of ?? =?? +?? and ?? -?? = ?? ?? -?????? ??? +?????? ??? ???????? ??? -?????? ??? , find ?? (?? ) subject to the condition, ?? ( ?? ?? )= ?? -?? ?? . (2011 : 12 Marks) Solution: Let ??? (?? )=?? +???? ??????? (?? )=???? -?? ???(1+??)?? (?? )=?? -?? +??(?? +?? ) ?=?? +???? ????? =?? -?? = ?? ?? -cos??? +sin??? cosh??? -cos??? ?= cosh??? +sin?h?? -cos??? +sin??? cos?h?? -cos??? ?=1+ sin?h?? +sin??? cos?h?? -cos??? ??? ??? =?? 1 (?? ,?? ) and ??? ??? =?? 2 (?? ,?? ) ????? 1 (?? ,?? )= ??? ??? = cos??? (cosh??? -cos??? )-sin??? (sinh??? +sin??? ) (cosh??? -cos??? ) 2 ????? 1 (?? ,0)= cos??? (1-cos??? )-sin 2 ??? (1-cos??? ) 2 = cos??? -1 (1-cos??? ) 2 ?= -1 1-cos??? = -1 2 cosec 2 ? ?? 2 ?? 2 (?? ,?? )= ??? ??? = cosh??? (cosh??? -cos??? )-sinh??? (sinh??? +sin??? ) (cosh??? ?cos??? ) 2 ?? 2 (?? ,0)= 1-cos??? (1-cos??? ) 2 = 1 1-cos?? = 1 2 cosec 2 ? ?? 2 ? By Mine's Method (1+??)?? (?? )?=??[?? 1 (?? ,0)-???? 2 (?? ,0)]???? +?? ?=??(- 1 2 cosec 2 ? ?? 2 -?? 1 2 cosec 2 ? ?? 2 )???? +?? ?=- 1 2 (1+??)??cosec 2 ? ?? 2 ???? +?? ?=(1+??)cot? ?? 2 +?? ???? (?? )?=cot? ?? 2 + ?? 1+?? ?=cot? ?? 2 +?? ,?? = ?? 1+?? At ?? = ?? 2 ,??? (?? )?= 3-?? 2 ???? ?=?? ( ?? 2 )-cot? ?? 4 = 1-?? 2 ???? (?? )?=cot? ?? 2 + 1-?? 2Read More
|
387 videos|203 docs
|
Related Exams
About this Document
|
Oct 31, 2024 Last updated |
Document Description: Analytic Functions for UPSC 2024 is part of Mathematics Optional Notes for UPSC preparation.
The notes and questions for Analytic Functions have been prepared according to the UPSC exam syllabus. Information about Analytic Functions covers topics
like and Analytic Functions Example, for UPSC 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Analytic Functions.
Introduction of Analytic Functions in English is available as part of our Mathematics Optional Notes for UPSC
for UPSC & Analytic Functions in Hindi for Mathematics Optional Notes for UPSC course.
Download more important topics related with notes, lectures and mock test series for UPSC
Exam by signing up for free. UPSC: Analytic Functions | Mathematics Optional Notes for UPSC
Description
Full syllabus notes, lecture & questions for Analytic Functions | Mathematics Optional Notes for UPSC - UPSC | Plus excerises question with solution to help you revise complete syllabus for Mathematics Optional Notes for UPSC | Best notes, free PDF download
Information about Analytic Functions
In this doc you can find the meaning of Analytic Functions defined & explained in the simplest way possible. Besides explaining types of
Analytic Functions theory, EduRev gives you an ample number of questions to practice Analytic Functions tests, examples and also practice UPSC
tests
|
Explore Courses for UPSC exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Analytic Functions | Mathematics Optional Notes for UPSC
,shortcuts and tricks
,ppt
,Important questions
,Previous Year Questions with Solutions
,study material
,past year papers
,Viva Questions
,video lectures
,practice quizzes
,Analytic Functions | Mathematics Optional Notes for UPSC
,mock tests for examination
,Extra Questions
,Free
,Exam
,Summary
,Objective type Questions
,Sample Paper
,Analytic Functions | Mathematics Optional Notes for UPSC
,Semester Notes
,MCQs
;
Additional Information about Analytic Functions for UPSC Preparation
Analytic Functions Free PDF Download
The Analytic Functions is an invaluable resource that delves deep into the core of the UPSC exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Analytic Functions now and kickstart your journey towards success in the UPSC exam.
Importance of Analytic Functions
The importance of Analytic Functions cannot be overstated, especially for UPSC aspirants.
This document holds the key to success in the UPSC exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Analytic Functions Notes
Analytic Functions Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Analytic Functions.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Analytic Functions Notes on EduRev are your ultimate resource for success.
Analytic Functions UPSC Questions
The "Analytic Functions UPSC Questions" guide is a valuable resource for all aspiring students preparing for the
UPSC exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Analytic Functions on the App
Students of UPSC can study Analytic Functions alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Analytic Functions,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Analytic Functions is prepared as per the latest UPSC syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup