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We have to prove the following identity:

$$z_1 \bar{z_2} = \frac{1}{4}(|z_1 + z_2|^2 + i|z_1 + iz_2|^2 - |z_1 - z_2|^2 - i|z_1 - iz_2|^2)$$

It says to use the identity we just proved, which is $\Re(z_1 \bar{z_2}) = \frac{1}{4}(|z_1 + z_2|^2 - |z_1 - z_2|^2)$

I get that we consider the identity we have proven with $\bar{z_2}$ and also with substituting $\bar z_2 \to i \bar z_2$. We also use the fact that $z = \Re(z) + i \Im(z)$, but I'm confused to the signs: Since $\Re(iz) = - \Im (z)$, shouldn't the terms multiplied by $i$ have the opposite sign than they do in the identity?

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  • $\begingroup$ Here since modulus is taken. Sign does not matter actually. $\endgroup$ – Ri-Li Sep 23 '14 at 18:35
  • $\begingroup$ That's just $\Re(z_1 \bar z_2) + i \Re(z_1 \bar z_2)$..? $\endgroup$ – MCT Sep 23 '14 at 18:36
  • $\begingroup$ No $\Re (z_1\bar z_2) + i \Im (z_1\bar z_2)$ $\endgroup$ – Ri-Li Sep 23 '14 at 18:42
  • $\begingroup$ The sign does matter since I'm talking about the sign outside of the modulus brackets. And placing an "i" outside of the real part of a number does not magically make that real part the imaginary part. ex $z = x+iy$, you gave me $x+ix$ not $x+iy$. $\endgroup$ – MCT Sep 23 '14 at 18:45
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Note that

$$\Im\{z_1\bar{z_2}\}=\Re\{-iz_1\bar{z_2}\}=\Re\{z_1\overline{iz_2}\}=\frac14(|z_1+iz_2|^2-|z_1-iz_2|^2)$$

So the identity in your question is actually correct.

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  • $\begingroup$ Oh right, I didn't account for the fact that all of $iz_2$ was under the bar sign. Thanks. +1 $\endgroup$ – MCT Sep 23 '14 at 18:51

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