Complex number basic stuff The first question is to prove that $\operatorname{Re}(iz)=-\operatorname{Im}(z)$.
First $z= x+ iy$;
$$\operatorname{Re}(i(x+iy))=\operatorname{Re}(ix+i^2y)=\operatorname{Re}(ix-y)$$
---> I don't know how to move on to make it equal to $-\operatorname{Im}(z)$.
Another similar question is to prove that $\operatorname{Im}(iz)=\operatorname{Re}(z)$.
Basically, $$\operatorname{Im}(iz)=\operatorname{Im}(ix-y)$$ 
---> same place I got stuck.
I know that $\operatorname{Re}(z)=x$, and $\operatorname{Im}(z)=y$.
 A: Remember what the $Re$ function does; takes a complex number as its input and gives back the real part. Similarly done for $Im$. So it follows then that $$Re(ix-y)=-y.$$ Very similar method for the $Im$ side; Let $z=x+iy$, then $$-Im(z)=-y.$$
A: Remember that $z+\bar z = 2\operatorname{Re}(z)$ and $z-\bar z = 2i\operatorname{Im}(z)$, and that $\overline{ab} = \bar a \, \bar b$. Then:
$$
2\operatorname{Re}(iz) = iz+\overline{iz} = iz + \bar i \, \bar z = iz - i \bar z = i (z-\bar z)=2i \cdot i(z-\bar z)=2i^2(z-\bar z) = -2 \operatorname{Im}(z)
$$
A: Note that $x,y$ are real. For the complex number, $-y+ix$, the real part is $-y$ and the imaginary part is $x$.
$$\Re(ix-y)=\Re(-y+ix)=-y=-\Im(z)$$
$$\Im(ix-y)=\Im(-y+ix)=x=\Re(z)$$
A: for any two complex numbers $a$ and $b$ we have (by definition of complex multiplication):
$$
\mathfrak{Re}(ab) = \mathfrak{Re}(a)\mathfrak{Re}(b) -\mathfrak{Im}(a)\mathfrak{Im}(b)
$$
setting $a=i$ gives $\mathfrak{Re}(a)=0$ and $\mathfrak{Im}(a)=1$, from which your result follows
