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Home » Find the square root of the following complex numbers. (i) 1 – i (ii) – 8 – 6i
Find the square root of the following complex numbers.
(i) 1 – i
(ii) – 8 – 6i
CBSE | Class 11 | Complex Numbers | Exercise 13.3 | Maths | RD Sharma
Find the square root of the following complex numbers.
(i) 1 – i
(ii) – 8 – 6i

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) 1-\mathrm{i}
It is known that, Z=(1-i)
Therefore, \sqrt{(a+i b)}=\sqrt{(1-i)}
Here, b<0
Let’s simplify now,
\begin{aligned} \sqrt{1-i} &=\pm\left[\left(\frac{1+\sqrt{(1)^{2}+(-1)^{2}}}{2}\right)^{\frac{1}{2}}-i\left(\frac{-1+\sqrt{(1)^{2}+(-1)^{2}}}{2}\right)^{\frac{1}{2}}\right] \\ &=\pm\left[\left(\frac{1+\sqrt{1+1}}{2}\right)^{\frac{1}{2}}-i\left(\frac{-1+\sqrt{1+1}}{2}\right)^{\frac{1}{2}}\right] \\ &=\pm\left[\left(\frac{1+\sqrt{2}}{2}\right)^{\frac{1}{2}}-i\left(\frac{-1+\sqrt{2}}{2}\right)^{\frac{1}{2}}\right] \\ &=\pm\left[\left(\sqrt{\frac{\sqrt{2}+1}{2}}\right)-i\left(\sqrt{\frac{\sqrt{2}-1}{2}}\right)\right] \end{aligned}
As a result, square root of (1-i) is \pm[(\sqrt{(\sqrt{2}+1) / 2)-i}(\sqrt{(\sqrt{2}-1) / 2)]}

(ii) -8 – 6
It is known that, Z=-8-6 \mathrm{i}
Therefore, \sqrt{(a+i b)}=-8-6 \mathrm{i}
Here, b<0
Let’s simplify now,
\begin{array}{l} \sqrt{-8-6 i}=\pm\left[\left(\frac{-8+\sqrt{(-8)^{2}+(-6)^{2}}}{2}\right)^{\frac{1}{2}}-i\left(\frac{8+\sqrt{(-8)^{2}+(-6)^{2}}}{2}\right)^{\frac{1}{2}}\right]_{[\text {Since, } b<0]} \\ =\pm\left[\left(\frac{-8+\sqrt{64+36}}{2}\right)^{\frac{1}{2}}-i\left(\frac{8+\sqrt{64+36}}{2}\right)^{\frac{1}{2}}\right] \\ =\pm\left[\left(\frac{-8+\sqrt{100}}{2}\right)^{\frac{1}{2}}-i\left(\frac{8+\sqrt{100}}{2}\right)^{\frac{1}{2}}\right] \\ =\pm\left[\left(\frac{-8+10}{2}\right)^{\frac{1}{2}}-i\left(\frac{8+10}{2}\right)^{\frac{1}{2}}\right] \\ \left.=\pm \pm\left(\frac{2}{2}\right)^{\frac{1}{2}}-i\left(\frac{18}{2}\right)^{\frac{1}{2}}\right] \\ =\left[1^{1 / 2}-i 9^{1 / 2}\right] \\ =\pm[1-3 i] \end{array}
As a result, square root of (-8-6 i) is \pm[1-3 i]

i

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