Question

let Z1 and z2 be two complex numbers then proof that |z1z2| = |Z1| |z2|​

Answer 1

Step-by-step explanation:

Let $\tt\pink{{z}_{1}={x}_{1}+i{y}_{1}\:\:\:\&\:\:\:{z}_{2}={x}_{2}+i{y}_{2} }$

So,

$\mathrm{ z_{1}z_{2} =(x_{1} + i \: y_{1} )( x_{2} + i \: y_{2} ) }$

$\mathrm{ \implies z_{1}z_{2} =( x_{1} x_{2} + i x_{1} y_{2} + ix_{2} y_{1} + {i}^{2} y_{1} y_{2}) }$

$\mathrm{ \implies z_{1}z_{2} = \{x_{1} x_{2} + i (x_{1} y_{2} + x_{2} y_{1} )- y_{1} y_{2} \} }$

$\mathrm{ \implies z_{1}z_{2} = (x_{1} x_{2} - y_{1} y_{2})+ i (x_{1} y_{2} + x_{2} y_{1} ) }$

Now,

$\mathrm{ | z_{1}z_{2} | = \sqrt{ (x_{1} x_{2} - y_{1} y_{2})^{2} + (x_{1} y_{2} + x_{2} y_{1} ) ^{2} } }$

$\mathrm{ \implies | z_{1}z_{2} | = \sqrt{ x_{1}^{2} x_{2} ^{2} + y_{1}^{2} y_{2}^{2} - 2 x_{1} x_{2}y_{1} y_{2} + x_{1}^{2} y_{2}^{2} + x_{2} ^{2} y_{1}^{2} + 2 x_{1} x_{2}y_{1} y_{2} } }$

$\mathrm{ \implies | z_{1}z_{2} | = \sqrt{ x_{1}^{2} x_{2} ^{2} + y_{1}^{2} y_{2}^{2} + x_{1}^{2} y_{2}^{2} + x_{2} ^{2} y_{1}^{2}} }$

$\mathrm{ \implies | z_{1}z_{2} | = \sqrt{ x_{1}^{2} x_{2} ^{2} + x_{1}^{2} y_{2}^{2} + y_{1}^{2} y_{2}^{2} + x_{2} ^{2} y_{1}^{2}} }$

$\mathrm{ \implies | z_{1}z_{2} | = \sqrt{ x_{1}^{2} ( x_{2} ^{2} + y_{2}^{2}) + y_{1}^{2} ( y_{2}^{2} + x_{2} ^{2} )} }$

$\mathrm{ \implies | z_{1}z_{2} | = \sqrt{ (x_{1}^{2} + y_{1}^{2})( x_{2} ^{2} + y_{2}^{2}) } }$

$\mathrm{ \implies | z_{1}z_{2} | = \sqrt{ (x_{1}^{2} + y_{1}^{2})}. \sqrt{( x_{2} ^{2} + y_{2}^{2}) } }$

$\red{ \mathsf{ \implies | z_{1}z_{2} | = | z_{1} |.| z_{2} | } }$