If `za n dw`are two complex number such that `|z w|=1a n da rg(z)-a rg(w)=pi/2`, then show that ` z w=-idot`

If `za n dw`are two complex number such that `|z w|=1a n da rg(z)-a rg

1.	If `za n dw`are two complex number such that `\|z w\|=1a n da rg(z)-a rg(w)=pi/2`, then show that ` z w=-idot`
Answer» Let `z = r_(1) (cos theta_(1) + isin theta_(1)) "and" w = r_(2) (costheta_(2) + isin theta _(2)) ` Also, `\|zw\| = \|z\|\|w\| = r_(1) r_(2) = 1 ` `:. r_(1) r_(2) = 1 ` Futher, `arg (z) = theta _(1) and arg (w) = theta_(2)` But ` arg (z) - arg (w) = (pi)/(2)` `rarr theta_(1) - theta_(2) = (pi)/(2)` `rarr arg((z)/(w)) = (pi)/(2)` Now. to prove `barzw = - i` LHS = `barzw` ` = r_(1) (costheta _(1) - isintheta _(1)) r_(2) (costheta_(2) + isin theta_(2))` ` = r_(1)r_(2) [cos(theta _(1) - theta_(2))+isin (theta_(2)-theta_(1) ) ]` ` = r_(1)r_(2) [cos(-pi//2)+isin (-pi//2 ) ]` `1 [0 - i]` `-i = RHS " "Hence proved.`