Question

Find the values of $\sqrt{4+3\sqrt{20i}}$.
A. $\sqrt{5}+2i$
B. $3+\sqrt{5}i$
C. $1+\sqrt{5}i$
D. $\sqrt{5}i-2i$

Answer 1

Hint: In the above question we have to find the square root of $\left( 4+3\sqrt{20i} \right)$ which is a complex number. So we will suppose a complex number ${{z}^{2}}={{\left( x+yi \right)}^{2}}=4+3\sqrt{20i}$ and on comparing real parts and imaginary parts. We will find the required values.

Complete step-by-step answer:
We have been given to find the values of $\sqrt{4+3\sqrt{20i}}$.
We have to find the square root of $\left( 4+3\sqrt{20i} \right)$.
Let us suppose ${{z}^{2}}={{\left( x+yi \right)}^{2}}=4+3\sqrt{20i}$.
On using the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$, we get as follows:

& \Rightarrow {{x}^{2}}+2xyi+{{\left( yi \right)}^{2}}=4+3\sqrt{20i} \\\ & \Rightarrow {{x}^{2}}+\left( 2xy \right)i-{{y}^{2}}=4+3\sqrt{20i} \\\ \end{aligned}$$ Since $${{i}^{2}}=-1$$, on rearranging the terms, we get as follows: $$\Rightarrow \left( {{x}^{2}}-{{y}^{2}} \right)+\left( 2xy \right)i=4+3\sqrt{20i}$$ On comparing real parts and imaginary parts of the above equation, we get as follows: $$\begin{aligned} & {{x}^{2}}-{{y}^{2}}=4.....(1) \\\ & 2xy=3\sqrt{20}.....(2) \\\ \end{aligned}$$ Now, consider the modulus: $${{\left| z \right|}^{2}}=\left| {{z}^{2}} \right|$$. Since $$\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$$ for $$\left( x+iy \right)$$ $$\Rightarrow \left| {{z}^{2}} \right|={{x}^{2}}+{{y}^{2}}$$ For $$\sqrt{4+3\sqrt{20i}}$$ we have, $$\begin{aligned} & \left| {{z}^{2}} \right|=\sqrt{{{\left( 4 \right)}^{2}}+{{\left( 3\sqrt{20i} \right)}^{2}}}=\sqrt{16+180}=\sqrt{196} \\\ & \Rightarrow {{x}^{2}}+{{y}^{2}}=\sqrt{196} \\\ & \Rightarrow {{x}^{2}}+{{y}^{2}}=14.....(3) \\\ \end{aligned}$$ Solving (1) and (3) we get as follows: $$\begin{aligned} & {{x}^{2}}-{{y}^{2}}+{{x}^{2}}+{{y}^{2}}=4+14 \\\ & 2{{x}^{2}}=18 \\\ & {{x}^{2}}=\dfrac{18}{2} \\\ & {{x}^{2}}=9 \\\ \end{aligned}$$ Taking square root on both the sides of the equation, we get as follows: $$x=\pm 3$$ For x=3, substitute x=3 in equation (2) we get as follows: $$\begin{aligned} & 2\times 3\times y=3\sqrt{20} \\\ & \Rightarrow 2y=\sqrt{20} \\\ & \Rightarrow y=\dfrac{\sqrt{20}}{2} \\\ & \Rightarrow y=\sqrt{5} \\\ \end{aligned}$$ For x=3, substitute x=-3 in equation (2) we get as follows: $$\begin{aligned} & 2\times \left( -3 \right)\times y=3\sqrt{20} \\\ & \Rightarrow -2y=\sqrt{20} \\\ & \Rightarrow y=\dfrac{-1}{2}\times \sqrt{20} \\\ & \Rightarrow y=-\sqrt{5} \\\ \end{aligned}$$ Hence, $$x+iy=\pm \left( 3+\sqrt{5i} \right)$$ Therefore, the correct answer of the given question is option B. Note: Be careful while doing calculation and also take care of the sign while calculation. Also we can use the direct formula to find the square root of a complex number which is given by as follows: $$a\pm bi=\pm \sqrt{\left( {{a}^{2}}-{{b}^{2}} \right)\pm 2abi}$$.