exp(i*theta)=exp{2*pi*i*[theta/(2*pi)]}=[exp(2*pi*i)]^[theta/(2*pi)]=1^[theta/(2*pi)]=1?exp(i*theta)=exp{2*pi*i*[theta/(2*pi)]}=[exp(2*pi*i)]^[theta/(2*pi)]=1^[theta/(2*pi)];exp(2*pi*i)等于1是吧!那这个式子哪里出问题了呢?难道复数
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/17 09:44:19
![exp(i*theta)=exp{2*pi*i*[theta/(2*pi)]}=[exp(2*pi*i)]^[theta/(2*pi)]=1^[theta/(2*pi)]=1?exp(i*theta)=exp{2*pi*i*[theta/(2*pi)]}=[exp(2*pi*i)]^[theta/(2*pi)]=1^[theta/(2*pi)];exp(2*pi*i)等于1是吧!那这个式子哪里出问题了呢?难道复数](/uploads/image/z/12459606-6-6.jpg?t=exp%28i%2Atheta%29%3Dexp%7B2%2Api%2Ai%2A%5Btheta%2F%282%2Api%29%5D%7D%3D%5Bexp%282%2Api%2Ai%29%5D%5E%5Btheta%2F%282%2Api%29%5D%3D1%5E%5Btheta%2F%282%2Api%29%5D%3D1%3Fexp%28i%2Atheta%29%3Dexp%7B2%2Api%2Ai%2A%5Btheta%2F%282%2Api%29%5D%7D%3D%5Bexp%282%2Api%2Ai%29%5D%5E%5Btheta%2F%282%2Api%29%5D%3D1%5E%5Btheta%2F%282%2Api%29%5D%EF%BC%9Bexp%282%2Api%2Ai%29%E7%AD%89%E4%BA%8E1%E6%98%AF%E5%90%A7%21%E9%82%A3%E8%BF%99%E4%B8%AA%E5%BC%8F%E5%AD%90%E5%93%AA%E9%87%8C%E5%87%BA%E9%97%AE%E9%A2%98%E4%BA%86%E5%91%A2%3F%E9%9A%BE%E9%81%93%E5%A4%8D%E6%95%B0)
exp(i*theta)=exp{2*pi*i*[theta/(2*pi)]}=[exp(2*pi*i)]^[theta/(2*pi)]=1^[theta/(2*pi)]=1?exp(i*theta)=exp{2*pi*i*[theta/(2*pi)]}=[exp(2*pi*i)]^[theta/(2*pi)]=1^[theta/(2*pi)];exp(2*pi*i)等于1是吧!那这个式子哪里出问题了呢?难道复数
exp(i*theta)=exp{2*pi*i*[theta/(2*pi)]}=[exp(2*pi*i)]^[theta/(2*pi)]=1^[theta/(2*pi)]=1?
exp(i*theta)=exp{2*pi*i*[theta/(2*pi)]}=[exp(2*pi*i)]^[theta/(2*pi)]=1^[theta/(2*pi)];
exp(2*pi*i)等于1是吧!那这个式子哪里出问题了呢?难道复数运算里面指数不能随便放入括号里吗?
求教
exp(i*theta)=exp{2*pi*i*[theta/(2*pi)]}=[exp(2*pi*i)]^[theta/(2*pi)]=1^[theta/(2*pi)]=1?exp(i*theta)=exp{2*pi*i*[theta/(2*pi)]}=[exp(2*pi*i)]^[theta/(2*pi)]=1^[theta/(2*pi)];exp(2*pi*i)等于1是吧!那这个式子哪里出问题了呢?难道复数
注意在复数范围内,1的方根可不止一个!而1^[theta/(2*pi)]可看作1的2pi/theta次方根.若2pi/theta是有理数,设2pi/theta=m/n,则1^[theta/(2*pi)]有t=lcm(m,n)个结果:{exp{i*2kpi/(2pi/theta)}|0<=k<=t-1}=exp(i*k*theta)|0<=k<=t-1};若2pi/theta不是有理数,则1^[theta/(2*pi)]有无穷个结果,遍历整个单位圆.原式=1仅是其中一种情况:当且仅当k=0时.此外还有t-1种.exp(i*theta)是k=1时得结果,也仅有这个才是原式的解,其它t-1种情况,全是增根.