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Complex numbers

  1. Oct 22, 2012 #1
    Can someone please prove the formulas:

    The real part of z= 1/2(z+z*)
    And the imaginary part of z= 1/2i(z-z*)

    I can't understand why it is like this. Could someone please give me the proof?
     
  2. jcsd
  3. Oct 22, 2012 #2
    What are ##z## and ##z^*## in terms of their real and imaginary parts?
     
  4. Oct 22, 2012 #3

    I write [itex]\,\overline z\,[/itex] instead of your z*. Put

    $$z=x+iy\,\,,\,\,x,y\in \Bbb R\Longrightarrow \,\,Re(z)=x\,\,,\,\,Im(z)=y\Longrightarrow$$

    $$z+\overline z=x+iy+x-iy=2x\;\;,\;\;z-\overline z=x+iy-(z-iy)=2yi$$

    Now end the exercise.

    DonAntonio
     
  5. Oct 22, 2012 #4
    So 1/2 2x = re part = 1/2(z+z*)
    And 1/2i 2yi= Im part= 1/2i(z-z*)

    Thanks!
     
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