Frsnel 4, Exercises of Aeronautical Engineering

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Next note that 2 e * dzalong the line from 0 to R as R — ow tends to (1). im /4 2 Next note that i e * dz along the arc of the circle of radius R from R to Re as R — o goes to zero. im/4 Finally, note that J, e* dz along the line from Re to Ois emit ee’ dx (3) 0 Therefore, (1) plus (3) is 0, so we get © id g, _ 1-iv™ [ ee dz= 2 (4) Taking the imaginary part of (4) yields that “i = [ sin(2?)dz =,