Complex number basic stuff

Question

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 $\operatorname{Im}(z)=y$.

Answer 1

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 $$-Im(z)=-y.$$

Answer 2

Note that $x,y$ are real. For the complex number, $-y+ix$, the real part is $-y$ and the imaginary part is $x$.

$$\Re(ix-y)=\Re(-y+ix)=-y=-\Im(z)$$

$$\Im(ix-y)=\Im(-y+ix)=x=\Re(z)$$

Answer 3

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:

$$ 2\operatorname{Re}(iz) = iz+\overline{iz} = iz + \bar i \, \bar z = iz - i \bar z = i (z-\bar z)=2i \cdot i(z-\bar z)=2i^2(z-\bar z) = -2 \operatorname{Im}(z) $$

Answer 4

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