If z ne 0 is a complex number, then

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4-1.Complex numbers

easy

If $z \ne 0$ is a complex number, then

${\mathop{\rm Re}\nolimits} (z) = 0 \Rightarrow {\mathop{\rm Im}\nolimits} ({z^2}) = 0$

${\mathop{\rm Re}\nolimits} ({z^2}) = 0 \Rightarrow {\mathop{\rm Im}\nolimits} ({z^2}) = 0$

${\mathop{\rm Re}\nolimits} (z) = 0 \Rightarrow {\mathop{\rm Re}\nolimits} ({z^2}) = 0$

None of these

Solution

(a) If $z \ne 0$. Let $z = x + iy$ ==> ${z^2} = {x^2} – {y^2} + i(2xy)$
$Re(z)= 0$ ==> $x = 0$. Therefore ${\mathop{\rm Im}\nolimits} ({z^2}) = 2xy = 0$
Thus ${\mathop{\rm Re}\nolimits} (z) = 0 \Rightarrow {\mathop{\rm Im}\nolimits} ({z^2}) = 0$.

Standard 11

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