z and z′ ∈ C . show that: 1. zz′^(−) =z^− ×z′^(−) 2. ((z/(z′)))^(−) =(z^− /(z′^(−) ))


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Question Number 118780 by mathocean1 last updated on 19/Oct/20

$z and z^{'} \in C .$ $show that :$ $1 . \overset{―}{{zz}^{'}} = \bar{z} \times \overset{―}{z^{'}}$ $2 . \overset{―}{(\frac{z}{z^{'}})} = \frac{\bar{z}}{\overset{―}{z^{'}}}$
	Answered by MJS_new last updated on 19/Oct/20

	$z_{n} = r_{n} e^{i θ_{n}}$ ${\bar{z}}_{n} = r_{n} e^{- i θ_{n}}$ $z_{1} z_{2} = r_{1} r_{2} e^{i (θ_{1} + θ_{2})}$ $\overset{―}{z_{1} z_{2}} = r_{1} r_{2} e^{- i (θ_{1} + θ_{2})} = r_{1} e^{- i θ_{1}} \times r_{2} e^{- i θ_{2}} = \overset{―}{z_{1}} \times \overset{―}{z_{2}}$ $similar for division$
	Commented by mathocean1 last updated on 19/Oct/20

	$thanks$
	Answered by Olaf last updated on 19/Oct/20

	$z = a + i b and z^{'} = a^{'} + {i b}^{'}$ $\overline{z} = a - i b and \overline{z^{'}} = a^{'} - {i b}^{'}$ ${z z}^{'} = {a a}^{'} - {b b}^{'} + ({a b}^{'} + a^{'} b) i$ $\overset{__}{{z z}^{'}} = {a a}^{'} - {b b}^{'} - ({a b}^{'} + a^{'} b) i (1)$ $\overline{z} \overline{z^{'}} = (a - i b) (a^{'} - {i b}^{'})$ $\overline{z} \overline{z^{'}} = {a a}^{'} - {b b}^{'} - ({a b}^{'} + a^{'} b) (2)$ $With (1) and (2) : \overset{__}{{z z}^{'}} = \overline{z} \overline{z^{'}}$ $Same reasoning for question 2 .$
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