- #1

- 100

- 0

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?

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- Thread starter CrazyNeutrino
- Start date

- #1

- 100

- 0

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

- 834

- 2

What are ##z## and ##z^*## in terms of their real and imaginary parts?

- #3

- 606

- 1

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?

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

- #4

- 100

- 0

So 1/2 2x = re part = 1/2(z+z*)

And 1/2i 2yi= Im part= 1/2i(z-z*)

Thanks!

And 1/2i 2yi= Im part= 1/2i(z-z*)

Thanks!

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