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Home » Find the square root of the following complex numbers. (i) – 5 + 12i (ii) -7 – 24i
Find the square root of the following complex numbers.
(i) – 5 + 12i
(ii) -7 – 24i
CBSE | Class 11 | Complex Numbers | Exercise 13.3 | Maths | RD Sharma
Find the square root of the following complex numbers.
(i) – 5 + 12i
(ii) -7 – 24i

Solution:

If b>0, \sqrt{a+i b}=\pm\left[\left(\frac{a+\sqrt{a^{2}+b^{2}}}{2}\right)^{\frac{1}{2}}+i\left(\frac{-a+\sqrt{a^{2}+b^{2}}}{2}\right)^{\frac{1}{2}}\right]
If b<0, \sqrt{a+i b}=\pm\left[\left(\frac{a+\sqrt{a^{2}+b^{2}}}{2}\right)^{\frac{1}{2}}-i\left(\frac{-a+\sqrt{a^{2}+b^{2}}}{2}\right)^{\frac{1}{2}}\right]

(i) -5+12 \mathrm{i}
It is known that, Z=a+i b
Therefore, \sqrt{(a+i b)}=\sqrt{(-5+12 i)}
Here, b>0
Let’s simplify now,
\begin{aligned} \sqrt{-5+12 i} =\pm\left[\left(\frac{-5+\sqrt{(-5)^{2}+12^{2}}}{2}\right)^{\frac{1}{2}}+i\left(\frac{5+\sqrt{(-5)^{2}+12^{2}}}{2}\right)^{\frac{1}{2}}\right][\text { Since, } b>0] \\ =\pm\left[\left(\frac{-5+\sqrt{25+144}}{2}\right)^{\frac{1}{2}}+i\left(\frac{5+\sqrt{25+144}}{2}\right)^{\frac{1}{2}}\right] \\ =\pm\left[\left(\frac{-5+\sqrt{169}}{2}\right)^{\frac{1}{2}}+i\left(\frac{5+\sqrt{169}}{2}\right)^{\frac{1}{2}}\right] \\ =\pm\left[\left(\frac{-5+13}{2}\right)^{\frac{1}{2}}+i\left(\frac{5+13}{2}\right)^{\frac{1}{2}}\right] \\ =\pm\left[\left(\frac{8}{2}\right)^{\frac{1}{2}}+i\left(\frac{18}{2}\right)^{\frac{1}{2}}\right] \\ =\pm\left[4^{\frac{1}{2}}+i 9^{\frac{1}{2}}\right] \\ =\pm[2+3 i] \end{aligned}
As a result, square root of (-5+12 \mathrm{i}) is \pm[2+3 i]

(ii) -7-24 i
It is known that, Z=-7-24 i
Therefore, \sqrt{(a+i b)}=\sqrt{(-7-24 i)}
Here, \mathrm{b}<0
Let’s simplify now,
\begin{aligned} \sqrt{-7-24 i} =\pm\left[\left(\frac{-7+\sqrt{(-7)^{2}+(-24)^{2}}}{2}\right)^{\frac{1}{2}}-i\left(\frac{7+\sqrt{(-7)^{2}+(-24)^{2}}}{2}\right)^{\frac{1}{2}}\right]_{[\text {Since, } b<0]} \\ =\pm\left[\left(\frac{-7+\sqrt{49+576}}{2}\right)^{\frac{1}{2}}-i\left(\frac{7+\sqrt{49+576}}{2}\right)^{\frac{1}{2}}\right] \\ =\pm\left[\left(\frac{-7+\sqrt{625}}{2}\right)^{\frac{1}{2}}-i\left(\frac{7+\sqrt{625}}{2}\right)^{\frac{1}{2}}\right] \\ =\pm\left[\left(\frac{-7+25}{2}\right)^{\frac{1}{2}}-i\left(\frac{7+25}{2}\right)^{\frac{1}{2}}\right] \\ =\pm\left[\left(\frac{18}{2}\right)^{\frac{1}{2}}-i\left(\frac{32}{2}\right)^{\frac{1}{2}}\right] \\ =\pm\left[9^{1 / 2}-\mathrm{i} 16^{1 / 2}\right] \\ =\pm[3-4 \mathrm{i}] \end{aligned}
As a result, square root of (-7-24 i) is \pm[3-4 i]

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