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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
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