Proof of Complex Number Formulas for Real & Imaginary Parts

[CrazyNeutrino]

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?

 

[Muphrid]

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

 

[DonAntonio]

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?

I write $\,\overline z\,$ 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

 

[CrazyNeutrino]

So 1/2 2x = re part = 1/2(z+z*)
And 1/2i 2yi= I am part= 1/2i(z-z*)

Thanks!

 

[CellCoree]

I am happy to provide a proof for these complex number formulas.

First, let's define z as a complex number in the form z = a + bi, where a is the real part and bi is the imaginary part.

For the first formula, the real part of z is given by Re(z) = a. We can also write z* as a - bi, since the conjugate of a complex number is the same real part but with the opposite sign for the imaginary part.

Therefore, (z + z*) = (a + bi) + (a - bi) = 2a, and dividing by 2 gives us 1/2(z + z*) = a. This matches the definition of the real part of z, proving the formula.

For the second formula, the imaginary part of z is given by Im(z) = bi. Similarly, z* can be written as a - bi.

Substituting into the formula, we have 1/2i(z - z*) = 1/2i[(a + bi) - (a - bi)] = 1/2i[bi + bi] = bi. This also matches the definition of the imaginary part of z, proving the formula.

In summary, these formulas are derived from the definition of the real and imaginary parts of a complex number and the properties of complex conjugates. I hope this helps clarify the reasoning behind these formulas.