(a) $\langle x,y+z\rangle=\langle x,y\rangle+\langle x,z\rangle$.
(b) $\langle x,cy\rangle =\bar c\langle x,y\rangle$.
(c) $\langle x,0\rangle = \langle 0,x\rangle =0$.
(d) $\langle x,x\rangle=0$ iff $x=0$.
(e) If $\langle x,y\rangle=\langle x,z\rangle$ for all $x$ belongs to $V$, then $y=z$.
$PROOF$(a): $\Longrightarrow$ We have, $\langle x,y+z\rangle=\overline{\langle y+z,x\rangle} = \overline{\langle y,x\rangle} + \overline{\langle z,x\rangle} = \langle x,y\rangle+\langle x,z\rangle$.
By using the conjugation criteria for inner product space i've also proved (b).But,i'm not getting how to proceed for (c),(d) &(e).
I need some hint.
Thank you!
Hint: If $\langle x,y\rangle=\langle x,z\rangle$ for every $x \in V$, then we can say that $$ \langle x,y-z \rangle = 0 $$ (again, for every $x \in V$). Now, what happens if we let $x = y-z$?