Re (1 + i) ^2 3 - i = - Vidyadip

MCQ

${\mathop{\rm Re}\nolimits} \frac{{{{(1 + i)}^2}}}{{3 - i}}$ =

✓
$- 1/5$
B
$1/5$
C
$1/10$
D
$-1/10$

✓

Answer

Correct option: A.

$- 1/5$

a
(a) ${\mathop{\rm Re}\nolimits} \left[ {\frac{{{{(1 + i)}^2}}}{{3 - i}}} \right] = {\mathop{\rm Re}\nolimits} \left[ {\left( {\frac{{2i}}{{3 - i}}} \right)\,\,\left( {\frac{{3 + i}}{{3 + i}}} \right)} \right]$
${\mathop{\rm Re}\nolimits} \left[ {\frac{{{{(1 + i)}^2}}}{{3 - i}}} \right] = {\mathop{\rm Re}\nolimits} \left[ {\left( {\frac{{2i}}{{3 - i}}} \right)\left( {\frac{{3 + i}}{{3 + i}}} \right)} \right]$
$ = {\mathop{\rm Re}\nolimits} \left[ {\frac{{6i - 2}}{{9 + 1}}} \right] = {\mathop{\rm Re}\nolimits} \left[ { - \frac{2}{{10}} + \frac{6}{{10}}i} \right] = - \frac{1}{5}$.

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