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

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

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Solution

Let $z_{1} = a_{1} + i b_{1} a n d z_{2} = a_{2} + i b_{2}$

Then $R e (z_{1}) = a_{1}, R e (z_{2}) = a_{2}, I m (z_{1}) = b_{1}$

and $I m (z_{2}) = b_{2}$

Now $z_{1} z_{2}$ $= (a_{1} + i b_{1}) (a_{2} + i b_{2}) = a_{1} a_{2} + i b_{1} b_{2} + i a_{1} b_{1} + i^{2} b_{1} b_{2} = (a_{1} a_{2} - b_{1} b_{2}) + i (a_{1} b_{2} + a_{2} b_{1}) (∵ i^{2} = - 1)$

$∴ R e (z_{1} z_{2}) = a_{1} a_{2} - b_{1} b_{2}$

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

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For any two complex numbers z1 and z2 prove that Re(z1z2)=Re(z1)Re(z2)−Im(z1)Im(z2)

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