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Q)

If $z^2=(0,1)$ find $z$.

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A)
Toolbox:
  • If $z=a+ib$ then $\bar{z}=a-ib$
  • $\mid z\mid=\sqrt{a^2+b^2}$
  • $z^{-1}=\large\frac{a-ib}{a^2+b^2}$
  • $z\bar{z}=a^2+b^2$
  • Also $Re(z)=a$,$Im(z)=b$
  • 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:
$z^2=(0,1)=i$
Therefore $z=\sqrt i=x+iy$
Squaring both sides we get
$i=(x^2-y^2)+2xyi$
Equating the real and imaginary parts separately
$x^2-y^2=0$-----(1)
$2xy=1$-----(2)
$2xy>0\Rightarrow x$ and $y$ are of the same sign.
Step 2:
Now $x^2+y^2=\sqrt{(x^2-y^2)^2+4x^2y^2}$
$x^2+y^2=1$----(3)
Solving (1) and (3)
$x^2=y^2=\large\frac{1}{2}$
$x=y=\pm\large\frac{1}{\sqrt 2}$
Therefore the roots are $\large\frac{1}{\sqrt 2}+\frac{i}{\sqrt 2},\frac{-1}{\sqrt 2}-\frac{i}{\sqrt 2}$
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