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4-1.Complex numbers
normal
If $z$ is a complex number, then which of the following is not true
A
$|{z^2}|\, = \,|z{|^2}$
B
$|{z^2}|\, = \,|\bar z{|^2}$
C
$z = \bar z$
D
${\bar z^2} = {\bar z^2}$
Solution
(c)$L.H.S.$= $|{z^2}|\, = \,|{(x + iy)^2}|$
$ = \,\,|{x^2} – {y^2} + 2ixy| = \sqrt {{{({x^2} – {y^2})}^2} + {{(2xy)}^2}} $
$ = \sqrt {{{\left( {{x^2} + {y^2}} \right)}^2}} $…..$(i)$
$R.H.S.$ $ = |z{|^2} = |x + iy{|^2} = \sqrt {{{({x^2} + {y^2})}^2}} $
$ = {x^2} + {y^2}$ ……$(ii)$
Therefore $|{z^2}| = |z{|^2}$
$(b)$ True $ (c)$ False (since $z \ne \overline z $ ).
Standard 11
Mathematics
