# Question #61d86

Aug 27, 2016

For A and B, let ${w}_{1} = {x}_{1} + i {y}_{1} , {w}_{2} = {x}_{2} + i {y}_{2}$, where ${x}_{1} , {x}_{2} , {y}_{1} , {y}_{2} \in \mathbb{R}$.

A) Proof:
$R e \left(i \cdot {w}_{1}\right) = R e \left(i \left({x}_{1} + i {y}_{1}\right)\right) = R e \left(- {y}_{1} + i {x}_{1}\right) = - {y}_{1} = - I m \left({x}_{1} + i {y}_{1}\right) = - I m \left({w}_{1}\right)$

B) Proof:
$I m \left({w}_{1} \cdot i\right) = I m \left(\left({x}_{1} + i {y}_{1}\right) i\right) = I m \left(- {y}_{1} + i {x}_{1}\right) = {x}_{1} = R e \left({x}_{1} + i {y}_{1}\right) = R e \left({w}_{1}\right)$

C) Counterexample: Let ${w}_{1} = 1 + i , {w}_{2} = 1 - i$. Then
$R e \left({w}_{1}\right) \cdot R e \left({w}_{2}\right) = 1 \cdot 1 = 1$, but
$R e \left({w}_{1} \cdot {w}_{2}\right) = R e \left(2\right) = 2$

D) Counterexample: Let ${w}_{1} = 1 , {w}_{2} = i$. Then
$\frac{I m \left({w}_{1}\right)}{I m \left({w}_{2}\right)} = \frac{0}{1} = 0$, but
$I m \left({w}_{1} / {w}_{2}\right) = I m \left(\frac{1}{i}\right) = I m \left(\frac{i}{i} ^ 2\right) = I m \left(- i\right) = - 1$