Question

設(shè)a是不等于1的正實數(shù),z=x+yi(x、y∈R).規(guī)定z^*=a^x(cosy+isiny),?

(1)求證:(z+2πi)^*=z^*;?

(2)設(shè)z₁、z₂為兩個復(fù)數(shù),試證明(z₁+z₂)^*=z₁^*·z₂^*;?

(3)類比(2)的結(jié)論寫出(z₁-z₂)^*的有關(guān)運(yùn)算式子,并證明你的結(jié)論.

Answer 1

證明:(1)∵z+2πi=x+(y+2π)i,?

∴(z+2πi)^*=a^x［cos(y+2π)+isin(y+2π)］??

=a^x(cosy+isiny)=z^*得證.?

(2)設(shè)z₁=x₁+y₁i,z₂=x₂+y₂i,z₁+z₂=x₁+x₂+(y₁+y₂)i.??

(z₁+z₂)^*=a^x1x2［cos(y₁+y₂)+isin(y₁+y₂）］,?

z₁^*·z₂^*=a^x1(cosy₁+isiny₁)·a^x2(cosy₂+isiny₂)?

=a^x1+x2［cos(y₁+y₂)+isin(y₁+y₂）］.?

∴（z₁+z₂)^*=z₁^*·z₂^*.?

(3)類比得(z₁-z₂)^*=.?

(z₁-z₂)^*=a^x1-x2［cos(y₁-y₂)+isin(y₁-y₂)］,?

=a^x1-x2 

=a^x1-x2［cos(y₁-y₂)+isin(y₁-y₂)］,得證.