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对于一个复数方程,两边取模会导致增解,而两边同时取共轭得到的是与原方程同解的方程,怎么会导致增解呢?但这样的奇怪事情却发生了:请看下面两例. 例1 已知z是复数,且z~3=z,求z. 解法一:在z~3=z两边取模得|z|~3=|z|,即|z|=1或|z|=0.若|z|=1,则在z~3=两边同乘以z得z~4=1,z=±1或z=±ι.连同z=0共五个解,代入原方程知都是原方程的解. 解法二:z~3=. ①两边同取共轭得=z ②把①中的=z~3代入到②式中得z~9=z,解得 z=0或z~8=1. 显然比上面解法多出4个根.奇怪的是①式与②式互为充要条件,是同解的,由它们联立的方程组所得的结果应该是它们的公共解,而解为什么能多呢?我们再看一例.
For a complex equation, modulo on both sides will lead to a solution, while the conjugates on both sides will result in an equation that shares the same solution as the original equation. How can this result in a solution? However, such a strange thing happened: Look at the following two Example 1. Known that z is a complex number, and z~3=z, find z. Solution one: Take the model on both sides of z~3=z |z|~3=|z|, ie, |z|=1 or |z|=0. If |z|=1, then multiply z by z=3=z by z=4=1, z=±1 or z=±ι. With z=0, there are five solutions. Into the original equation knows the solution of the original equation. Solution 2: z~3=. 1 Both sides take the same conjugate =z 2 Substitute =z~3 in 1 into 2 Where z~9=z, solution To get z=0 or z~8=1. Obviously there are 4 more roots than the above solution. It is curious that 1 and 2 are necessary and sufficient conditions for each other, and they are the same solution, and the result of their simultaneous equations It should be their public solution, and why can the solution be more? Let’s look at another example.