Question

Find the square root of the following complex numbers :

(i) -5 + 12 i

(ii) -7 - 24i

(iii) 1 - i

(iv) - 8 - 6i

(v) 8 - 15i

(vi) -11 - 64 $\sqrt{-1}$

(vii) 1 + 4 $\sqrt{-3}$

(viii) 4i

(ix) - i

Answer 1

$\left(i\right)-5+12i\text{Let z}=-5+12i\Rightarrow |z|=\sqrt{(-5{)}^2+{12}^2}=\sqrt{25+144}=\sqrt{169}=13\therefore \sqrt{-5+12i}=\pm \{\sqrt{\frac{13+(-5)}{2}}+i\sqrt{\frac{13-(-5)}{2}}\}=\pm \{\sqrt{\frac{8}{2}}+i\sqrt{\frac{18}{2}}\}(\because y>0)=\pm \{2+3i\}$

$\left(ii\right)-7-24i\text{Let}z=-7-24i\text{then}|z|=\sqrt{(-7{)}^2+(-24{)}^2}=\sqrt{49+576}=\sqrt{625}=25\therefore \sqrt{-7-24i}=\pm \{\sqrt{\frac{25-7}{2}}-i\sqrt{\frac{25+7}{2}}\}(\because y<0)=\pm (\sqrt{\frac{\sqrt{2}+1}{2}}-i\sqrt{\frac{\sqrt{2}-1}{2}})$

$\left(iii\right)1-i\text{Let}z=1-i\text{then}|z|=\sqrt{1^2+(-1{)}^2}=\sqrt{1+1}=\sqrt{2}\therefore \sqrt{1-i}=\pm (\sqrt{\frac{\sqrt{2}+1}{2}}-i\sqrt{\frac{\sqrt{2}-1}{2}})(\because y<0)=\pm (\sqrt{\frac{\sqrt{2}+1}{2}}-i\sqrt{\frac{\sqrt{2}-1}{2}})$

$\left(iv\right)-8-6i\text{Let}z=-8-6i\text{then}|z|=\sqrt{(-8{)}^2+(-6{)}^2}=\sqrt{64+36}=\sqrt{100}\therefore \sqrt{-8-6i}=\pm \{\sqrt{\frac{10-8}{2}}-i\sqrt{\frac{10+8}{2}}\}(\because y<0)=\pm \{\sqrt{\frac{2}{2}}-i\sqrt{\frac{18}{2}}\}=\pm \{\sqrt{1}-i\sqrt{9}\}=\pm \{1-3i\}$

$\left(v\right)8-15i\text{Let}z=8-15i\text{then}|z|=\sqrt{(8{)}^2+(-15{)}^2}=\sqrt{64+225}=\sqrt{289}=17\therefore \sqrt{8-5i}=\pm \{\sqrt{\frac{17+8}{2}}-i\sqrt{\frac{17-8}{2}}\}(\because y<0)=\pm \{\sqrt{\frac{25}{2}}-i\sqrt{\frac{9}{2}}\}=\pm \{\frac{5}{\sqrt{2}}-i\frac{3}{\sqrt{2}}\}=\pm \frac{1}{\sqrt{2}}\{5-3i\}$

$\left(vi\right)-11-64\sqrt{-1}\text{Let}z=-11-64\sqrt{-1}\Rightarrow z=-11-60i(\because \sqrt{-1}=i)\text{Then}|z|=\sqrt{(-11{)}^2+(-60{)}^2}=\sqrt{121+3600}=\sqrt{3721}=61\therefore \sqrt{-11-60i}=\pm \{\sqrt{\frac{61-11}{2}}-i\sqrt{\frac{61+11}{2}}\}(\because y<0)=\pm \{\sqrt{\frac{50}{2}}-i\sqrt{\frac{72}{2}}\}=\pm \{\sqrt{25}-i\sqrt{36}\}=\pm \{5-6i\}$

$\left(vii\right)1+4\sqrt{-3}\text{Let}z=1+4\sqrt{-3}=1+4\sqrt{3}\times \sqrt{-1}(\because \sqrt{-3}=\sqrt{3}\times \sqrt{-1})\Rightarrow z=1+4\sqrt{3i}\therefore |z|=\sqrt{(1{)}^2+(4\sqrt{3}{)}^2}=\sqrt{1+48}=\sqrt{49}=7\text{Hence}\sqrt{1+4\sqrt{-3}}=\pm \{\sqrt{\frac{7+1}{2}}+i\sqrt{\frac{7-1}{2}}\}(\because y>0)=\pm \{\sqrt{\frac{8}{2}}+i\sqrt{\frac{6}{2}}\}=\pm \{\sqrt{4}+i\sqrt{3}\}=\pm \{2+\sqrt{3}i\}$

$\left(viii\right)4i\text{Let}z=4i\text{then}|z|=4|i|=4|i|(\because |z_1{z}_2|=|z_1|\times |z_2|)=4(\because |i|=1)\therefore \sqrt{4}i=\pm \{\sqrt{\frac{4+0}{2}}+i\sqrt{\frac{4-0}{2}}\}(\because y>0)=\{\sqrt{2}+i\sqrt{2}\}=\pm \sqrt{2}(1+i)$

$\left(ix\right)-i\text{Let}z=-i\text{then}|z|=|-i|=|-1|\times |i|=1\times i(\because |z_1{z}_2|=|z_1|\times |z_2|)=1(\because |i|=1)\therefore \sqrt{-i}=\pm \{\sqrt{\frac{1+0}{2}}-1\sqrt{\frac{1-0}{2}}\}=\pm (\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}})=\pm \frac{1}{\sqrt{2}}(1-i)$