Question

prove that:-
${( z_{1} - z_{2} )}^{2} = { z_{1} }^{2} - 2 z_{1}z_{2} + { z_{2} }^{2}$
​

Answer 1

$\fbox{ \fbox{ \bold{ \large{ \: \: To \: Prove </p><p> \: \: }}}}$

$\to\bold{ {( z_{1} - z_{2} )}^{2} = { z_{1} }^{2} - 2 z_{1}z_{2} + { z_{2} }^{2} } \\ \:$

$\fbox{ \fbox{\bold{ \large{ \: Proof \: }}}}$

$\to\bold{ {( z_{1} - z_{2} )}^{2}} \: \\ \to\bold{(z_{1} -z_{2} )(z_{1} -z_{2} ) } \\ \to \bold{ z_{1}(z_{1} - z_{2}) - z_{2}(z_{1} - z_{2})} \\ \to \bold{\bold{ {(z_{1} )}^{2} } - z_{1}z_{2} - z_{1}z_{2} + {(z_{1})}^{2} } \\ \to \bold{ {(z_{1})}^{2} - 2z_{1}z_{2} + {(z_{2} )}^{2} }$

$\bold{ \underline{ \underline{ { \: \: \: Hence \: Proved \: \: \: }}}}$

Answer 2

Answer:

Pls refer to the image.

Hope it helps..pls mark as braineliest ☺️