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# For any two complex numbers $\;z_{1}\;$ and $\;z_{2}\;$, what is the value of $\; Re(z_{1} z_{2})$?

$(a)\;0\qquad(b)\;Rez_{1}Rez_{2} - Imz_{1}Imz_{2}\qquad(c)\;Rez_{1}Imz_{2} - Imz_{1}Rez_{2}\qquad(d)\;Imz_{1}Rez_{2} - Rez_{1}Imz_{2}$

Can you answer this question?

Answer : $\;Rez_{1}Rez_{2} - Imz_{1}Imz_{2}$
Explanation :
Let $\;z_{1}=x_{1}+iy_{1}\;$ and $\;z_{2} = x_{2}+iy_{2}$
Therefore , $\;z_{1}z_{2} = (x_{1}+iy_{1})(x_{2}+iy_{2})$
$=x_{1}(x_{2}+iy_{2}) + iy_{1} (x_{2}+iy_{2})$
$= x_{1}x_{2} +x_{1}iy_{2} +iy_{1}x_{2}+i^{2}y_{1}y_{2}$
$= x_{1}x_{2} +ix_{1}y_{2} +ix_{2}y_{1}-y_{1}y_{2}$
$= (x_{1}x_{2} - y_{1}y_{2}) + i (x_{1}y_{2}+x_{2}y_{1})$
$Re(z_{1}z_{2}) = x_{1}x_{2}-y_{1}y_{2}$
$\; Re(z_{1} z_{2})=Re z_{1} Rez_{2} - Im z_{1} Im z_{2}$
Hence proved .
answered Apr 15, 2014 by