\displaystyle \textbf{Question 1: }\text{Find the square roots of the following complex numbers:}
\displaystyle \text{(i) }-5+12i\qquad\text{(ii) }-7-24i\qquad\text{(iii) }1-i\qquad\text{(iv) }-8-6i
\displaystyle \text{(v) }8-15i\qquad\text{(vi) }-11-60\sqrt{-1}\qquad\text{(vii) }1+4\sqrt{-3}
\displaystyle \text{(viii) }4i\qquad\text{(ix) }-i
\displaystyle \text{Answer:}

There are two ways by which we can solve the problems. We have a few solved by each method.

\displaystyle \text{(i) Let }\sqrt{-5+12i}=x+iy.
\displaystyle \text{Squaring both sides,}
\displaystyle -5+12i=(x+iy)^2
\displaystyle \Rightarrow -5+12i=(x^2-y^2)+2xyi
\displaystyle \Rightarrow x^2-y^2=-5\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \Rightarrow 2xy=12\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle \text{Now, }(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2
\displaystyle \Rightarrow (x^2+y^2)^2=(-5)^2+12^2
\displaystyle \Rightarrow (x^2+y^2)^2=169
\displaystyle \Rightarrow x^2+y^2=13\qquad\ldots\ldots\ldots\text{(iii)}
\displaystyle \left[\because x^2+y^2>0\right]
\displaystyle \text{Adding equations (i) and (iii),}
\displaystyle 2x^2=8
\displaystyle \Rightarrow x^2=4
\displaystyle \Rightarrow x=\pm2
\displaystyle \text{From equation (i),}
\displaystyle y^2=x^2+5
\displaystyle \Rightarrow y^2=4+5=9
\displaystyle \Rightarrow y=\pm3
\displaystyle \text{Since }2xy=12>0,\text{ therefore }x\text{ and }y\text{ have the same signs.}
\displaystyle \therefore \sqrt{-5+12i}=\pm(2+3i).

\displaystyle \text{(ii) }\sqrt{z}=\pm\left\{\sqrt{\frac{|z|+\text{Re}(z)}{2}}+i\sqrt{\frac{|z|-\text{Re}(z)}{2}}\right\},\text{ if }\text{Im}(z)>0
\displaystyle \sqrt{z}=\pm\left\{\sqrt{\frac{|z|+\text{Re}(z)}{2}}-i\sqrt{\frac{|z|-\text{Re}(z)}{2}}\right\},\text{ if }\text{Im}(z)<0
\displaystyle \text{Let }z=-7-24i.
\displaystyle \therefore \text{Re}(z)=-7
\displaystyle |z|=\sqrt{(-7)^2+(-24)^2}=\sqrt{49+576}=25
\displaystyle \text{Here, }\text{Im}(z)<0.
\displaystyle \therefore \sqrt{-7-24i}=\pm\left\{\sqrt{\frac{25+(-7)}{2}}-i\sqrt{\frac{25-(-7)}{2}}\right\}
\displaystyle =\pm\left\{\sqrt{\frac{18}{2}}-i\sqrt{\frac{32}{2}}\right\}
\displaystyle =\pm\left(\sqrt{9}-i\sqrt{16}\right)
\displaystyle \therefore \sqrt{-7-24i}=\pm(3-4i).

\displaystyle \text{(iii) }\sqrt{z}=\pm\left\{\sqrt{\frac{|z|+\text{Re}(z)}{2}}+i\sqrt{\frac{|z|-\text{Re}(z)}{2}}\right\},\text{ if }\text{Im}(z)>0
\displaystyle \sqrt{z}=\pm\left\{\sqrt{\frac{|z|+\text{Re}(z)}{2}}-i\sqrt{\frac{|z|-\text{Re}(z)}{2}}\right\},\text{ if }\text{Im}(z)<0
\displaystyle \text{Let }z=1-i.
\displaystyle \therefore \text{Re}(z)=1
\displaystyle |z|=\sqrt{1^2+(-1)^2}=\sqrt{2}
\displaystyle \text{Here, }\text{Im}(z)<0.
\displaystyle \therefore \sqrt{1-i}=\pm\left\{\sqrt{\frac{\sqrt{2}+1}{2}}-i\sqrt{\frac{\sqrt{2}-1}{2}}\right\}.

\displaystyle \text{(iv) }\sqrt{z}=\pm\left\{\sqrt{\frac{|z|+\text{Re}(z)}{2}}+i\sqrt{\frac{|z|-\text{Re}(z)}{2}}\right\},\text{ if }\text{Im}(z)>0
\displaystyle \sqrt{z}=\pm\left\{\sqrt{\frac{|z|+\text{Re}(z)}{2}}-i\sqrt{\frac{|z|-\text{Re}(z)}{2}}\right\},\text{ if }\text{Im}(z)<0
\displaystyle \text{Let }z=-8-6i.
\displaystyle \therefore \text{Re}(z)=-8
\displaystyle |z|=\sqrt{(-8)^2+(-6)^2}=\sqrt{64+36}=10
\displaystyle \text{Here, }\text{Im}(z)<0.
\displaystyle \therefore \sqrt{-8-6i}=\pm\left\{\sqrt{\frac{10-8}{2}}-i\sqrt{\frac{10+8}{2}}\right\}
\displaystyle =\pm\left(\sqrt{1}-i\sqrt{9}\right)
\displaystyle \therefore \sqrt{-8-6i}=\pm(1-3i).

\displaystyle \text{(v) }\sqrt{z}=\pm\left\{\sqrt{\frac{|z|+\text{Re}(z)}{2}}+i\sqrt{\frac{|z|-\text{Re}(z)}{2}}\right\},\text{ if }\text{Im}(z)>0
\displaystyle \sqrt{z}=\pm\left\{\sqrt{\frac{|z|+\text{Re}(z)}{2}}-i\sqrt{\frac{|z|-\text{Re}(z)}{2}}\right\},\text{ if }\text{Im}(z)<0
\displaystyle \text{Let }z=8-15i.
\displaystyle \therefore \text{Re}(z)=8
\displaystyle |z|=\sqrt{8^2+(-15)^2}=\sqrt{64+225}=17
\displaystyle \text{Here, }\text{Im}(z)<0.
\displaystyle \therefore \sqrt{8-15i}=\pm\left\{\sqrt{\frac{17+8}{2}}-i\sqrt{\frac{17-8}{2}}\right\}
\displaystyle =\pm\left(\frac{5}{\sqrt{2}}-\frac{3}{\sqrt{2}}i\right)
\displaystyle \therefore \sqrt{8-15i}=\pm\frac{1}{\sqrt{2}}(5-3i).

\displaystyle \text{(vi) Let }\sqrt{-11-60i}=x+iy.
\displaystyle \text{Squaring both sides,}
\displaystyle -11-60i=(x+iy)^2
\displaystyle \Rightarrow -11-60i=(x^2-y^2)+2xyi
\displaystyle \Rightarrow x^2-y^2=-11\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \Rightarrow 2xy=-60\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle \text{Now, }(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2
\displaystyle \Rightarrow (x^2+y^2)^2=(-11)^2+(-60)^2
\displaystyle \Rightarrow (x^2+y^2)^2=121+3600=3721
\displaystyle \Rightarrow x^2+y^2=61\qquad\ldots\ldots\ldots\text{(iii)}
\displaystyle \left[\because x^2+y^2>0\right]
\displaystyle \text{Adding equations (i) and (iii),}
\displaystyle 2x^2=50
\displaystyle \Rightarrow x^2=25
\displaystyle \Rightarrow x=\pm5
\displaystyle \text{From equation (i),}
\displaystyle y^2=x^2+11
\displaystyle \Rightarrow y^2=25+11=36
\displaystyle \Rightarrow y=\pm6
\displaystyle \text{Since }2xy=-60<0,\text{ therefore }x\text{ and }y\text{ have opposite signs.}
\displaystyle \therefore \sqrt{-11-60i}=\pm(5-6i).

\displaystyle \text{(vii) Let }\sqrt{1+4\sqrt{3}i}=x+iy.
\displaystyle \text{Squaring both sides,}
\displaystyle 1+4\sqrt{3}i=(x+iy)^2
\displaystyle \Rightarrow 1+4\sqrt{3}i=(x^2-y^2)+2xyi
\displaystyle \Rightarrow x^2-y^2=1\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \Rightarrow 2xy=4\sqrt{3}\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle \text{Now, }(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2
\displaystyle \Rightarrow (x^2+y^2)^2=1^2+(4\sqrt{3})^2
\displaystyle \Rightarrow (x^2+y^2)^2=1+48=49
\displaystyle \Rightarrow x^2+y^2=7\qquad\ldots\ldots\ldots\text{(iii)}
\displaystyle \left[\because x^2+y^2>0\right]
\displaystyle \text{Adding equations (i) and (iii),}
\displaystyle 2x^2=8
\displaystyle \Rightarrow x^2=4
\displaystyle \Rightarrow x=\pm2
\displaystyle \text{From equation (i),}
\displaystyle y^2=x^2-1
\displaystyle \Rightarrow y^2=4-1=3
\displaystyle \Rightarrow y=\pm\sqrt{3}
\displaystyle \text{Since }2xy=4\sqrt{3}>0,\text{ therefore }x\text{ and }y\text{ have the same signs.}
\displaystyle \therefore \sqrt{1+4\sqrt{3}i}=\pm(2+\sqrt{3}i).

\displaystyle \text{(viii) }\sqrt{z}=\pm\left\{\sqrt{\frac{|z|+\text{Re}(z)}{2}}+i\sqrt{\frac{|z|-\text{Re}(z)}{2}}\right\},\text{ if }\text{Im}(z)>0
\displaystyle \sqrt{z}=\pm\left\{\sqrt{\frac{|z|+\text{Re}(z)}{2}}-i\sqrt{\frac{|z|-\text{Re}(z)}{2}}\right\},\text{ if }\text{Im}(z)<0
\displaystyle \text{Let }z=4i.
\displaystyle \therefore \text{Re}(z)=0
\displaystyle |z|=\sqrt{0^2+4^2}=4
\displaystyle \text{Here, }\text{Im}(z)>0.
\displaystyle \therefore \sqrt{4i}=\pm\left\{\sqrt{\frac{4+0}{2}}+i\sqrt{\frac{4-0}{2}}\right\}
\displaystyle =\pm\left(\sqrt{2}+i\sqrt{2}\right)
\displaystyle \therefore \sqrt{4i}=\pm\sqrt{2}(1+i).

\displaystyle \text{(ix) }\sqrt{z}=\pm\left\{\sqrt{\frac{|z|+\text{Re}(z)}{2}}+i\sqrt{\frac{|z|-\text{Re}(z)}{2}}\right\},\text{ if }\text{Im}(z)>0
\displaystyle \sqrt{z}=\pm\left\{\sqrt{\frac{|z|+\text{Re}(z)}{2}}-i\sqrt{\frac{|z|-\text{Re}(z)}{2}}\right\},\text{ if }\text{Im}(z)<0
\displaystyle \text{Let }z=-i.
\displaystyle \therefore \text{Re}(z)=0
\displaystyle |z|=\sqrt{0^2+(-1)^2}=1
\displaystyle \text{Here, }\text{Im}(z)<0.
\displaystyle \therefore \sqrt{-i}=\pm\left\{\sqrt{\frac{1+0}{2}}-i\sqrt{\frac{1-0}{2}}\right\}
\displaystyle =\pm\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)
\displaystyle \therefore \sqrt{-i}=\pm\frac{1}{\sqrt{2}}(1-i).

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