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Just a question about complex vs real linearity. The problem is to show that $T: \mathbb{C} \rightarrow \mathbb{C}$ is $\mathbb{C}$-linear if and only if $T(iz) = iTz$ for all z.

So far, I understand that some relation T is $\mathbb{R}$-linear if $T(m \vec{v} \pm n \vec{u}) = mT(\vec{v}) \pm nT(\vec{u})$ for $m, n \in \mathbb{R}$ and it’s $\mathbb{C}$-linear if the same thing is true for $m, n \in \mathbb{C}$.

I’m not really sure where to go from here. I found $iz = -z_2 + iz_1$, but I’m not sure how to use the definitions of linearity here. Thanks in advance.

Edited to fix awful LaTeX, sorry

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    $\begingroup$ What is $CC$, what is $RR$, what is $T$? $\endgroup$ – Thorgott Jan 18 '20 at 22:54
  • $\begingroup$ Use \mathbb{C} for $\mathbb{C}$, \vec{v} for $\vec{v}$ and \in for $\in$. $\endgroup$ – azif00 Jan 18 '20 at 22:55
  • $\begingroup$ @Azif00: thank you $\endgroup$ – Sam Jan 18 '20 at 23:00
  • $\begingroup$ What you found (one line before the last one) doesn't make much sense if you don't specify what $\;z,\,z_1,\,z_2\;$ are, since that equality is clearly false for general $\;z,z_1,z_2\in\Bbb C\;$ $\endgroup$ – DonAntonio Jan 18 '20 at 23:02
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A simple counter-example is $T(z)=|z|$. Unless you assume that $T$ is already real linear you cannot prove this.

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