# Show that $T: \mathbb{C} \rightarrow \mathbb{C}$ is $\mathbb{C}$-linear if and only if $T(iz) = iTz$ for all z

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

• What is $CC$, what is $RR$, what is $T$? – Thorgott Jan 18 '20 at 22:54
• Use \mathbb{C} for $\mathbb{C}$, \vec{v} for $\vec{v}$ and \in for $\in$. – azif00 Jan 18 '20 at 22:55
• @Azif00: thank you – Sam Jan 18 '20 at 23:00
• 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 $\;z,z_1,z_2\in\Bbb C\;$ – DonAntonio Jan 18 '20 at 23:02

A simple counter-example is $$T(z)=|z|$$. Unless you assume that $$T$$ is already real linear you cannot prove this.