If |{z_1} + {z_2}| = |{z_1} - {z_2}|, then th | vedclass

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Question

(c) Squaring the given relations implies that
${x_1}{x_2} + {y_1}{y_2} = 0$
Now $amp\,\,{z_1} - amp\,\,{z_2} = {	an ^{ - 1}}\frac{{{y_1}}}{{{x_1}}}

Vedclass · Accepted Answer

$\frac{\pi }{2}$

Vedclass · Answer

(c) Squaring the given relations implies that
${x_1}{x_2} + {y_1}{y_2} = 0$
Now $amp\,\,{z_1} - amp\,\,{z_2} = {	an ^{ - 1}}\frac{{{y_1}}}{{{x_1}}} - {	an ^{ - 1}}\frac{{{y_2}}}{{{x_2}}}$
$ = {	an ^{ - 1}}\frac{{\frac{{{y_1}}}{{{x_1}}} - \frac{{{y_2}}}{{{x_2}}}}}{{1 + \frac{{{y_1}{y_2}}}{{{x_1}{x_2}}}}} = {	an ^{ - 1}}\frac{{{y_1}{x_2} - {y_2}{x_1}}}{{{x_1}{x_2} + {y_1}{y_2}}}$$ = {	an ^{ - 1}}\infty = \frac{\pi }{2}$.

If $|{z_1} + {z_2}| = |{z_1} - {z_2}|$, then the difference in the amplitudes of ${z_1}$ and ${z_2}$ is

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