CBSE BoardEnglish MediumSTD 11 ScienceMathsModel Paper 13 Marks
Question
Find the square root of $3 - 4i$
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Answer
Let $x + yi =\sqrt{3-4 i}$
Squaring both sides, we get
$x^2-y^2+2 x y i=3-4 i$
Equating the real and imaginary parts
$x^2-y^2=3 \ldots . \text { (i) }$
and $2 x y=-4$
$\Rightarrow x y=-2$
Now from the identity, we know
$\left(x^2+y^2\right)=\left(x^2-y^2\right)^2+4 x^2 y^2$
$\left(x^2+y^2\right)^2=(3)^2+4(-2)^2$
$=9+16=25$
$\therefore x^2+y^2=5 \ldots (ii) [$Neglecting $(-)$ sign as $\left.x^2+y^2>0\right]$
Solving $(i)$ and $(ii)$ we get
$ x ^2=4$ and $y ^2=1$
$x= \pm 2$ and $y= \pm 1$
Since the sign of $xy$ is negative
$\therefore$ if, $x =2, y =-1$
and if $x =-2, y =1$
$\therefore \sqrt{-5+12 i}= \pm(2-i)$
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