For any two complex numbers $z_{1}$ and $z_{2}$ , prove that $R e (z_{1} z_{2}) = R e z_{1} R e z_{2} - I m z_{1} I m z_{2}$

Open in App

Solution

Let $z_{1} = x_{1} + i y_{1}$ and $z_{2} = x_{2} + i y_{2}$
$\Rightarrow z_{1} z_{2} = (x_{1} + {i y}_{1}) (x_{2} + {i y}_{2})$
$= x_{1} x_{2} + i (y_{1} x_{2} + x_{1} y_{2}) + i^{2} y_{1} y_{2}$
$= (x_{1} x_{2} - y_{1} y_{2}) + i (y_{1} x_{2} + x_{1} y_{2})$ , since $(i^{2} = - 1)$
Hence $R e (z_{1} z_{2}) = R e (z_{1}) R e (z_{2}) - I m (z_{1}) I m (z_{2})$

Suggest Corrections

Similar questions

For any two complex numbers $z_{1}$ and $z_{2}$ prove that

$R e (z_{1} z_{2}) = R e (z_{1}) R e (z_{2}) - I m (z_{1}) I m (z_{2})$

z_{1}

z_{2}

R e (z_{1} z_{2}) = R e z_{1} R e z_{2} - I m z_{1} I m z_{2}

z_{1}, z_{2}

| z_{1} + z_{2} |^{2} = | z_{1} |^{2} + | z_{2} |^{2} .

z_{1}

z_{2}

| z_{1} | + | z_{2} | = ∣ ∣ ∣ \frac{z_{1} + z_{2}}{2} + \sqrt{z_{1} z_{2}} ∣ ∣ ∣ + ∣ ∣ ∣ \frac{z_{1} + z_{2}}{2} - \sqrt{z_{1} z_{2}} ∣ ∣ ∣

z_{1}, z_{2}

(| z_{1} | + | z_{2} |) ∣ ∣ ∣ \frac{z_{1}}{| z_{1} |} + \frac{z_{2}}{| z_{2} |} ∣ ∣ ∣ \leq 2 (| z_{1} | + | z_{2} |)

Join BYJU'S Learning Program