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Q)

Find the square root of $(-8-6i)$

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A)
Toolbox:
  • If $a+ib=c+id$ then $a=c$ and $b=d$
  • (i.e) if $z_1=z_2$,then $Re(z_1)=Re(z_2),Im(z_1)=Im(z_2)$
Step 1:
Let $\sqrt{-8-6i}=x+iy$
Squaring both sides,
$-8-6i=(x+iy)^2$
$-8-6i=x^2-y^2+2xyi$
Equating the real and imaginary parts separately
$x^2-y^2=-8$----(1)
$2xy=-6$----(2)
Step 2:
Since $2xy <0\Rightarrow x,y$ are of opposite signs.
Now $x^2+y^2=\sqrt{(x^2-y^2)^2+4x^2y^2}$
$\qquad\qquad\;\;\;\;=\sqrt{(-8)^2+(-6)^2}$
$\qquad\qquad\;\;\;\;=\sqrt{64+36}$
$\qquad\qquad\;\;\;\;=\sqrt{100}$
$x^2+y^2=10$-----(3)
Step 3:
Solving (1) and (3) we get
$x^2=1$,$y^2=9$ or $x=\pm 1,y=\pm 3$
Since $x$ and $y$ are of opposite signs,the roots are $1-3i,-1+3i$
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