For any two complex numbers {z_1},{z_2}we hav | vedclass

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Question

(a) We have $|{z_1} + {z_2}{|^2} = |{z_1}{|^2} + |{z_2}{|^2}$
==> $|{z_1}{|^2} + |{z_2}{|^2} + 2|{z_1}||{z_2}|\cos ({	heta _1} - {	heta _2})$
$

Vedclass · Accepted Answer

${\mathop{m Re}
olimits} \left( {\frac{{{z_1}}}{{{z_2}}}} ight) = 0$

Vedclass · Answer

(a) We have $|{z_1} + {z_2}{|^2} = |{z_1}{|^2} + |{z_2}{|^2}$
==> $|{z_1}{|^2} + |{z_2}{|^2} + 2|{z_1}||{z_2}|\cos ({	heta _1} - {	heta _2})$
$ = |{z_1}{|^2} + |{z_2}{|^2}$
Where ${	heta _1} = arg({z_1}),{	heta _2} = arg({z_2})$
==> $\cos ({	heta _1} - {	heta _2}) = 0\,\,\,\, \Rightarrow {	heta _1} - {	heta _2} = \frac{\pi }{2}$
==> $arg\left( {\frac{{{z_1}}}{{{z_2}}}} ight) = \frac{\pi }{2} \Rightarrow {\mathop{m Re}
olimits} \left( {\frac{{{z_1}}}{{{z_2}}}} ight) = \frac{{|{z_1}|}}{{|{z_2}|}}\cos \left( {\frac{\pi }{2}} ight) = 0$
Note : Also ${\mathop{m Re}
olimits} \left( {\frac{{{z_1}}}{{{z_2}}}} ight) = 0 \Rightarrow {\mathop{m Re}
olimits} ({z_1}\overline {{z_2}} ) = 0$
==> ${z_1}\overline {{z_2}} $ is purely imaginary.

For any two complex numbers ${z_1},{z_2}$we have $|{z_1} + {z_2}{|^2} = $ $|{z_1}{|^2} + |{z_2}{|^2}$ then

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