→ akrsw :C: (0,0) 到 (2,4) 的直線 → y = 2x, 0 <= x <= 2 03/11 10:04
→ akrsw :x + i2y = x + i4x = (1+4i)x 03/11 10:04
→ akrsw :dz = dx + idy = dx + 2idx = (1+2i)dx 03/11 10:04
→ akrsw :原積分 =∫(1+4i)(1+2i)xdx, 0 <= x <= 2 03/11 10:05
→ akrsw : |2 03/11 10:05
→ akrsw := (-7+6i)x^2/2| = (-7+6i)*2 = -14+12i 03/11 10:05
→ akrsw : |0 03/11 10:05
→ polarfox :謝謝a大解答..我想請問一下說z = x + iy 跟積分內 03/11 23:14
→ polarfox :x + i2y 是沒有關係的嗎? 03/11 23:14