If {z_1},{z_2} in C, then amp,left( {frac{{{{ | vedclass

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Question

(c) $arg\left( {\frac{{{z_1}}}{{{{\bar z}_2}}}} \right) = arg\,{z_1} - arg\,({\overline z _2}) = arg\,{z_1} + arg\,{z_2}$$ = arg({z_1}.{z_2})$
Option

Vedclass · Accepted Answer

$amp\,\left( {\frac{{{z_2}}}{{{{\bar z}_1}}}} \right)$

Vedclass · Answer

(c) $arg\left( {\frac{{{z_1}}}{{{{\bar z}_2}}}} \right) = arg\,{z_1} - arg\,({\overline z _2}) = arg\,{z_1} + arg\,{z_2}$$ = arg({z_1}.{z_2})$
Option $(c) $ gives the same result.

If ${z_1},{z_2} \in C$, then $amp\,\left( {\frac{{{{\rm{z}}_{\rm{1}}}}}{{{{{\rm{\bar z}}}_{\rm{2}}}}}} \right) = $

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