Find the square root of the following complex numbers. (i) -i contexto.140 3 years ago Solution: If If (i) It is known that, Therefore, Here, Let’s simplify now, As a result, square root of is
![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]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5ffa871c6cd7675f669d73da69f0953d_l3.png "Rendered by QuickLaTeX.com")
![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]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-bc2b1e8b97142f16b93977b37fecd267_l3.png "Rendered by QuickLaTeX.com")
![\begin{aligned} \sqrt{-i} &=\pm\left[\left(\frac{0+\sqrt{0^{2}+(-1)^{2}}}{2}\right)^{\frac{1}{2}}-i\left(\frac{0+\sqrt{0^{2}+(-1)^{2}}}{2}\right)^{\frac{1}{2}}\right]_{[\text {Since, } b<0]} \\ &=\pm\left[\left(\frac{0+\sqrt{0+1}}{2}\right)^{\frac{1}{2}}-i\left(\frac{0+\sqrt{0+1}}{2}\right)^{\frac{1}{2}}\right] \\ &=\pm\left[\left(\frac{0+\sqrt{1}}{2}\right)^{\frac{1}{2}}-i\left(\frac{0+\sqrt{1}}{2}\right)^{\frac{1}{2}}\right] \\ &=\pm\left[\left(\frac{0+1}{2}\right)^{\frac{1}{2}}-i\left(\frac{0+1}{2}\right)^{\frac{1}{2}}\right] \\ &=\pm\left[\left(\frac{1}{2}\right)^{\frac{1}{2}}-i\left(\frac{1}{2}\right)^{\frac{1}{2}}\right] \\ &=\pm\left[\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right] \\ &=\pm 1 / \sqrt{2}(1-\mathrm{i}) \end{aligned}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3b28e4c07af291b4f50988e1cd54767d_l3.png "Rendered by QuickLaTeX.com")
