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Combined Elementary transformation : LINEAR TRANSFORMATION : ๐ค = ๐ผ๐ง+๐ฝ, Example:- ๐ค = (1+๐)๐ง+(2-๐) .
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Complex Analysis Theorem from Conformal transformation
(Combined Elementary transformation )
๐ณ๐ฐ๐ต๐ฌ๐จ๐น ๐ป๐น๐จ๐ต๐บ๐ญ๐ถ๐น๐ด๐จ๐ป๐ฐ๐ถ๐ต :-
The transformation of the form
๐ค = ๐ผ๐ง+๐ฝ
Where ๐ผ and ๐ฝ both are complex constant then this transformation is known as a linear transformation .
This transformation is the resultant of magnification, rotation and translation.
๐ฌ๐ฟ๐จ๐ด๐ท๐ณ๐ฌ :-
Consider a linear transformation
๐ค = (1+๐)๐ง+(2-๐) and determine the region in ๐น' in ๐ค plane into which the corresponding rectangular region ๐น bounded by ๐ฅ=0, ๐ฆ=0, ๐ฅ=1 and ๐ฆ=2 and in ๐ง plane.
๐๐๐๐๐๐๐๐ :-
Given that
๐ค = (1+๐)๐ง+(2-๐)
๐ข+๐๐ฃ=(1+๐)(๐ฅ+๐๐ฆ)+(2-๐)
๐ข+๐๐ฃ=(๐ฅ+๐๐ฅ+๐๐ฆ-๐ฆ+2-๐)
๐ข+๐๐ฃ=(๐ฅ-๐ฆ+2)+๐(๐ฅ+๐ฆ-1)
๐ข=(๐ฅ-๐ฆ+2) and ๐ฃ=(๐ฅ+๐ฆ-1)
โต the rectangular region ๐น bounded by the lines ๐ฅ=0, ๐ฆ=0, ๐ฅ=2 and ๐ฆ=3 and in ๐ง plane.
at ๐ฅ=0 then ๐ข=-๐ฆ+2 and ๐ฃ= ๐ฆ-1 โ ๐ข+๐ฃ=1
โด ๐ข=0, ๐ฃ=1 โ (0,1) and ๐ข=1, ๐ฃ=0 โ (1,0)
at ๐ฆ=0 then ๐ข=๐ฅ+2 and ๐ฃ=๐ฅ-1 โ ๐ข-๐ฃ=3
โด ๐ข=0, ๐ฃ=-3 โ (0,-3) and ๐ข=3, ๐ฃ=0 โ (3,0)
at ๐ฅ=1 then ๐ข=-๐ฆ+3 and ๐ฃ=๐ฆ โ ๐ข+๐ฃ=3
โด ๐ข=0, ๐ฃ=3 โ (0,3) and ๐ข=3, ๐ฃ=0 โ (3,0)
at y=2 then ๐ข=๐ฅ and ๐ฃ=๐ฅ+1 โ ๐ข-๐ฃ=-1
โด ๐ข=0, ๐ฃ=1 โ (0,1) and ๐ข=-1, ๐ฃ=0 โ (-1,0)
Hence
the required image is a rectangle region ๐น' bounded by the lines ๐ข+๐ฃ=1, ๐ข-๐ฃ=3 , ๐ข+๐ฃ=3 and ๐ข-๐ฃ=-1 in ๐ค plane corresponding to rectangular region ๐น bounded by ๐ฅ=0, ๐ฆ=0, ๐ฅ=1 and ๐ฆ=2 and in ๐ง plane.
here โต ๐ผ=(1+๐)=โ2.๐โฑโฝ โทซแโดโพ the figure in ๐ค plane is โ2 times, rotate (๐ /4) and ๐ฝ=(2-๐) origin shift (2,-1) and of figure of ๐ง plane .
See figure in video.
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Solved...!!!
.
.
.
#complexanalysis #csirnet #conformal
#isogonal #transformation #mapping #lineartransformation
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(Combined Elementary transformation )
๐ณ๐ฐ๐ต๐ฌ๐จ๐น ๐ป๐น๐จ๐ต๐บ๐ญ๐ถ๐น๐ด๐จ๐ป๐ฐ๐ถ๐ต :-
The transformation of the form
๐ค = ๐ผ๐ง+๐ฝ
Where ๐ผ and ๐ฝ both are complex constant then this transformation is known as a linear transformation .
This transformation is the resultant of magnification, rotation and translation.
๐ฌ๐ฟ๐จ๐ด๐ท๐ณ๐ฌ :-
Consider a linear transformation
๐ค = (1+๐)๐ง+(2-๐) and determine the region in ๐น' in ๐ค plane into which the corresponding rectangular region ๐น bounded by ๐ฅ=0, ๐ฆ=0, ๐ฅ=1 and ๐ฆ=2 and in ๐ง plane.
๐๐๐๐๐๐๐๐ :-
Given that
๐ค = (1+๐)๐ง+(2-๐)
๐ข+๐๐ฃ=(1+๐)(๐ฅ+๐๐ฆ)+(2-๐)
๐ข+๐๐ฃ=(๐ฅ+๐๐ฅ+๐๐ฆ-๐ฆ+2-๐)
๐ข+๐๐ฃ=(๐ฅ-๐ฆ+2)+๐(๐ฅ+๐ฆ-1)
๐ข=(๐ฅ-๐ฆ+2) and ๐ฃ=(๐ฅ+๐ฆ-1)
โต the rectangular region ๐น bounded by the lines ๐ฅ=0, ๐ฆ=0, ๐ฅ=2 and ๐ฆ=3 and in ๐ง plane.
at ๐ฅ=0 then ๐ข=-๐ฆ+2 and ๐ฃ= ๐ฆ-1 โ ๐ข+๐ฃ=1
โด ๐ข=0, ๐ฃ=1 โ (0,1) and ๐ข=1, ๐ฃ=0 โ (1,0)
at ๐ฆ=0 then ๐ข=๐ฅ+2 and ๐ฃ=๐ฅ-1 โ ๐ข-๐ฃ=3
โด ๐ข=0, ๐ฃ=-3 โ (0,-3) and ๐ข=3, ๐ฃ=0 โ (3,0)
at ๐ฅ=1 then ๐ข=-๐ฆ+3 and ๐ฃ=๐ฆ โ ๐ข+๐ฃ=3
โด ๐ข=0, ๐ฃ=3 โ (0,3) and ๐ข=3, ๐ฃ=0 โ (3,0)
at y=2 then ๐ข=๐ฅ and ๐ฃ=๐ฅ+1 โ ๐ข-๐ฃ=-1
โด ๐ข=0, ๐ฃ=1 โ (0,1) and ๐ข=-1, ๐ฃ=0 โ (-1,0)
Hence
the required image is a rectangle region ๐น' bounded by the lines ๐ข+๐ฃ=1, ๐ข-๐ฃ=3 , ๐ข+๐ฃ=3 and ๐ข-๐ฃ=-1 in ๐ค plane corresponding to rectangular region ๐น bounded by ๐ฅ=0, ๐ฆ=0, ๐ฅ=1 and ๐ฆ=2 and in ๐ง plane.
here โต ๐ผ=(1+๐)=โ2.๐โฑโฝ โทซแโดโพ the figure in ๐ค plane is โ2 times, rotate (๐ /4) and ๐ฝ=(2-๐) origin shift (2,-1) and of figure of ๐ง plane .
See figure in video.
------------------------------------------------------------------
Solved...!!!
.
.
.
#complexanalysis #csirnet #conformal
#isogonal #transformation #mapping #lineartransformation
.
๐ฅ๐ฅ๐ฅHurry Up Students for whole detailed Videos with Answer in Description for making notes ๐ฅ๐ฅ๐ฅ
๐ฅ๐ฅ๐ฅJOIN JOIN JOIN JOIN ๐ฅ๐ฅ๐ฅ
๐ฅ๐ฅ๐ฅAll B.Sc., B.Tech , M.Sc. , M.A. Mathematics students join with us A great platform to get knowledge in Mathematics and science... Stay tuned....for detailed videos on whole SYLLABUS with topic wise...and also join us for CSIR-UGC National Eligibility Test (NET) for Junior Research Fellowship and Lecturer-ship COMMON SYLLABUS FOR PART 'B' AND 'C' MATHEMATICAL SCIENCE ๐ฅ๐ฅ๐ฅ
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