例3指出命题的错误:z≠0,有Ln(-z)=Lmz 证∵(-z)2=z2 (1) Ln(z)=lna 2)Lnz≠2Lnz Ln(-z+ Ln(-z)=Lnz Lnz (3 Lnz+Lmz≠2Lnz 2Ln(-z)=2Lnz(4) . Ln( z)=ln (5) {6+2kx}+{6+2k} Lnz=In z|+i(8+ 2kn)=2in z +2i(8+2kx) Ln(z)=In z +i0+(2k+1)xt例 指出命题的错误 有 3 : 0, ( ) .   − = z Ln z Lnz 2 2 证 ( ) (1) − = z z 2 2  − = Ln z Lnz ( ) (2)  − + − = + Ln z Ln z Lnz Lnz ( ) ( ) (3)  − = 2 ( ) 2 (4) Ln z Lnz  − = Ln z Lnz ( ) (5) Ln z ln z i k ( ) | | [ (2 1) ] − = + + +   Lnz ln z i k = + + | | ( 2 )   2 Lnz Lnz  2 Lnz Lnz Lnz +  2  + + 2 | | 2 ( 2 ) ln z i k   { 2 }   + k + + { 2 }  k