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MODULAR FORMS-page71 - 67 Use the function q = e2i to map...

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67 Use the function q = e 2 πiτ to map the segment { τ : | Re τ | ≤ 1 2 , Im τ = h } onto the circle C : | q | = e - 2 πh . When we move along the segment from the point 1 2 + ih to the point - 1 2 + ih the image point moves along the circle in the clockwise way. We have 1 2 πi Z ∂P 1 f dτ f = - 1 2 πi Z C (2 πiν ( f ) q ν ( f ) + . . . ) dq 2 πiq ( a ν ( f ) q ν ( f ) + . . . ) = - ν ( f ) . If we integrate along the part ∂P 2 of the boundary of P which lies on the circle C r ( ρ 2 ) we get lim r 0 1 2 πi Z ∂P 2 f dτ f = - 1 6 ν ρ 2 ( f ) . This is because the arc ∂P 2 approaches to the one-sixth of the full circle when its radius goes to zero. Also we take into account that the direction of the path is clockwise. Similarly, if we let ∂P 3 = ∂P C r ( i ) , ∂P 4 = ∂P C r ( ρ ), we find lim r 0 1 2 πi Z ∂P 3 f dτ f = - 1 2 ν i ( f ) . lim r 0 1 2 πi Z ∂P 4 f dτ f = - 1 6 ν ρ ( f ) . Now the transformation T : τ τ + 1 transforms the path along ∂P from - 1 2 + ih to ρ 2 to the path along the boundary from the point ρ to the point 1 2 + ih .
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