Question

Find the square root of the following complex numbers:
(i) −5 + 12i
(ii) −7 − 24i
(iii) 1 − i
(iv) −8 − 6i
(v) 8 −15i
(vi) $-11-60\sqrt{-1}$
(vii) $1+4\sqrt{-3}$
(viii) 4i
(ix) −i

Answer 1

$$\begin{gathered} \sqrt{z}=\pm \left[\sqrt{\frac{\left|z\right|+\mathrm{Re}\left(z\right)}{2}}+i\sqrt{\frac{\left|z\right|-\mathrm{Re}\left(z\right)}{2}}\right],\mathrm{if}\mathrm{Im}\left(z\right)>0 \\ \sqrt{z}=\pm \left[\sqrt{\frac{\left|z\right|+\mathrm{Re}\left(z\right)}{2}}-i\sqrt{\frac{\left|z\right|-\mathrm{Re}\left(z\right)}{2}}\right],\mathrm{if}\mathrm{Im}\left(z\right)<0 \\ \left(\mathrm{i}\right)z=-5+12i,\mathrm{Re}\left(z\right)=-5\mathrm{and}\left|z\right|=\sqrt{25+144}=13 \\ \mathrm{Here},\mathrm{Im}\left(z\right)>0 \\ \therefore \sqrt{z}=\pm \left[\sqrt{\frac{\left|z\right|+\mathrm{Re}\left(z\right)}{2}}+i\sqrt{\frac{\left|z\right|-\mathrm{Re}\left(z\right)}{2}}\right] \\ =\pm \left[\sqrt{\frac{13-5}{2}}+i\sqrt{\frac{13+5}{2}}\right] \\ =\pm \left[\sqrt{4}+i\sqrt{9}\right] \\ =\pm \left(2+3i\right) \\ \left(\mathrm{ii}\right)z=-7-24i,\mathrm{Re}\left(z\right)=-7,\left|z\right|=\sqrt{49+576}=25 \\ \mathrm{Here},\mathrm{Im}\left(z\right)<0 \\ \sqrt{z}=\pm \left[\sqrt{\frac{\left|z\right|+\mathrm{Re}\left(z\right)}{2}}-i\sqrt{\frac{\left|z\right|-\mathrm{Re}\left(z\right)}{2}}\right] \\ =\pm \left[\sqrt{\frac{25-7}{2}}-i\sqrt{\frac{25+7}{2}}\right] \\ =\pm \left(3-4i\right) \\ \left(\mathrm{iii}\right)z=1-i,\mathrm{Re}\left(z\right)=1,\left|z\right|=\sqrt{1+1}=\sqrt{2} \\ \mathrm{Here},\mathrm{Im}\left(z\right)<0 \\ \therefore \sqrt{z}=\pm \left[\sqrt{\frac{\left|z\right|+\mathrm{Re}\left(z\right)}{2}}-i\sqrt{\frac{\left|z\right|-\mathrm{Re}\left(z\right)}{2}}\right] \\ =\pm \left[\sqrt{\frac{\sqrt{2}+1}{2}}-i\sqrt{\frac{\sqrt{2}-1}{2}}\right] \\ \left(\mathrm{iv}\right)z=-8-6i,\mathrm{Re}\left(z\right)=-8,\left|z\right|=\sqrt{64+36}=10 \\ \mathrm{Here},\mathrm{Im}\left(z\right)<0 \\ \therefore \sqrt{z}=\pm \left[\sqrt{\frac{\left|z\right|+\mathrm{Re}\left(z\right)}{2}}-i\sqrt{\frac{\left|z\right|-\mathrm{Re}\left(z\right)}{2}}\right] \\ =\pm \left[\sqrt{\frac{10-8}{2}}-i\sqrt{\frac{10+8}{2}}\right] \\ =\pm \left(1-3i\right) \\ \left(\mathrm{v}\right)z=8-15i,\mathrm{Re}\left(z\right)=8,\left|z\right|=\sqrt{64+225}=17 \\ \mathrm{Here},\mathrm{Im}\left(z\right)<0 \\ \therefore \sqrt{z}=\pm \left[\sqrt{\frac{\left|z\right|+\mathrm{Re}\left(z\right)}{2}}-i\sqrt{\frac{\left|z\right|-\mathrm{Re}\left(z\right)}{2}}\right] \\ =\pm \left[\sqrt{\frac{17+8}{2}}-i\sqrt{\frac{17+8}{2}}\right] \\ =\pm \frac{1}{\sqrt{2}}\left(5-3i\right) \\ \left(\mathrm{vi}\right)-11-60\sqrt{-1}=-11-60i,\mathrm{Re}\left(z\right)=-11,\left|z\right|=\sqrt{121+3600}=61 \\ \mathrm{Here},\mathrm{Im}\left(z\right)<0 \\ \therefore \sqrt{z}=\pm \left[\sqrt{\frac{\left|z\right|+\mathrm{Re}\left(z\right)}{2}}-i\sqrt{\frac{\left|z\right|-\mathrm{Re}\left(z\right)}{2}}\right] \\ =\pm \left[\sqrt{\frac{61-11}{2}}-i\sqrt{\frac{61+11}{2}}\right] \\ =\pm \left(5-6i\right) \\ \left(\mathrm{vii}\right)z=1+4\sqrt{3}\sqrt{-1}=1+4\sqrt{3}i,\mathrm{Re}\left(z\right)=1,\left|z\right|=\sqrt{1+16\times 3}=7 \\ \mathrm{Here},\mathrm{Im}\left(z\right)>0 \\ \therefore \sqrt{z}=\pm \left[\sqrt{\frac{\left|z\right|+\mathrm{Re}\left(z\right)}{2}}+i\sqrt{\frac{\left|z\right|-\mathrm{Re}\left(z\right)}{2}}\right] \\ =\pm \left[\sqrt{\frac{7+1}{2}}+i\sqrt{\frac{7-1}{2}}\right] \\ =\pm \left(2+\sqrt{3}i\right) \\ \left(\mathrm{viii}\right)z=0+4i,\mathrm{Re}\left(z\right)=0,\left|z\right|=4 \\ \mathrm{Here},\mathrm{Im}\left(z\right)>0 \\ \therefore \sqrt{z}=\pm \left[\sqrt{\frac{\left|z\right|+\mathrm{Re}\left(z\right)}{2}}+i\sqrt{\frac{\left|z\right|-\mathrm{Re}\left(z\right)}{2}}\right] \\ =\pm \left[\sqrt{\frac{4+0}{2}}+i\sqrt{\frac{4-0}{2}}\right] \\ =\pm \left(\sqrt{2}+i\sqrt{2}\right) \\ =\pm \sqrt{2}\left(1+i\right) \\ \left(\mathrm{ix}\right)z=-i,\mathrm{Re}\left(z\right)=0,\left|z\right|=1 \\ \mathrm{Here},\mathrm{Im}\left(z\right)<0 \\ \therefore \sqrt{z}=\pm \left[\sqrt{\frac{\left|z\right|+\mathrm{Re}\left(z\right)}{2}}-i\sqrt{\frac{\left|z\right|-\mathrm{Re}\left(z\right)}{2}}\right] \\ =\pm \left[\sqrt{\frac{1}{2}}-i\sqrt{\frac{1}{2}}\right] \\ =\pm \frac{1}{\sqrt{2}}\left(1-i\right) \end{gathered}$$