作业帮求两道复变函数极限【请用定义证明】：（1）lim(z -> z0 ) z^2 = z0^2（2）lim(z -> 1-i ) [x+i(2x+y)] = 1 + i ,其中z = x + iy
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![求两道复变函数极限【请用定义证明】：（1）lim(z -> z0 ) z^2 = z0^2（2）lim(z -> 1-i ) [x+i(2x+y)] = 1 + i ,其中z = x + iy](/uploads/image/z/7251380-44-0.jpg?t=%E6%B1%82%E4%B8%A4%E9%81%93%E5%A4%8D%E5%8F%98%E5%87%BD%E6%95%B0%E6%9E%81%E9%99%90%E3%80%90%E8%AF%B7%E7%94%A8%E5%AE%9A%E4%B9%89%E8%AF%81%E6%98%8E%E3%80%91%EF%BC%9A%EF%BC%881%EF%BC%89lim%28z+-%3E+z0+%29+z%5E2+%3D+z0%5E2%EF%BC%882%EF%BC%89lim%28z+-%3E+1-i+%29+%5Bx%2Bi%282x%2By%29%5D+%3D+1+%2B+i+%2C%E5%85%B6%E4%B8%ADz+%3D+x+%2B+iy)
求两道复变函数极限【请用定义证明】：（1）lim(z -> z0 ) z^2 = z0^2（2）lim(z -> 1-i ) [x+i(2x+y)] = 1 + i ,其中z = x + iy
求两道复变函数极限
【请用定义证明】：
（1）lim(z -> z0 ) z^2 = z0^2
（2）lim(z -> 1-i ) [x+i(2x+y)] = 1 + i ,其中z = x + iy
求两道复变函数极限【请用定义证明】：（1）lim(z -> z0 ) z^2 = z0^2（2）lim(z -> 1-i ) [x+i(2x+y)] = 1 + i ,其中z = x + iy
（1）对于∀ε>0,∃δ=min｛1,ε｝,当|z-z0|<δ时,
（2）对于∀ε>0,∃δ=ε,当|z-（1-i）|<δ时,有|x-1|<δ,|y+1|<δ,此时,