Question

设a是不等于1的正实数,z=x+yi(x、y∈R).规定z^*=a^x(cosy+isiny),?

(1)求证:(z+2πi)^*=z^*;?

(2)设z₁、z₂为两个复数,试证明(z₁+z₂)^*=z₁^*·z₂^*;?

(3)类比(2)的结论写出(z₁-z₂)^*的有关运算式子,并证明你的结论.

Answer 1

证明:(1)∵z+2πi=x+(y+2π)i,?

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

=a^x(cosy+isiny)=z^*得证.?

(2)设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)类比得(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)］,得证.