show that Norm of v+iT(v) equals to the Norm of v-iT(v)

Question

So I have $V$ an inner product space above $C$ and a linear operator $T$ such as $T=T^*$ on $V$

I need to prove that:

$$||v+iT(v)||=||v−iT(v)||$$

I tried to write it by definition, but I didn't get any useful. I also don't understand how to use the hermitian characteristic which given.

Answer 1

Some fun facts used below: $\langle u,tv\rangle=\bar{t}\langle u,v\rangle$, $\langle tu,v\rangle=t\langle u,v\rangle$, and $\langle T(u),v\rangle=\langle u,T^{*}(v)\rangle$, where $t \in \Bbb{C}$ is a scalar. \begin{align*} \|v+iT(v)\|^2&=\langle v+iT(v),\, v+iT(v)\rangle\\ &=\langle v,v\rangle + \langle v,iT(v)\rangle + \langle iT(v),v\rangle + \langle iT(v),iT(v)\rangle\\ &=\|v\|^2-i\langle v,T(v)\rangle+i\langle T(v),v\rangle+\|T(v)\|^2\\ &=\|v\|^2-i\langle v,T(v)\rangle+i\langle v,T^{*}(v)\rangle+\|T(v)\|^2 \\ &=\|v\|^2-i\langle v,T(v)\rangle+i\langle v,T(v)\rangle+\|T(v)\|^2 && (\because T^{*}=T)\\ &=\|v\|^2+\|T(v)\|^2. \end{align*} Now do the same with RHS to see the equality.