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:
\[\begin{align}
  & \Rightarrow {{x}^{2}}+2xyi+{{\left( yi \right)}^{2}}=4+3\sqrt{20i} \\
 & \Rightarrow {{x}^{2}}+\left( 2xy \right)i-{{y}^{2}}=4+3\sqrt{20i} \\
\end{align}\]
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:
\[\begin{align}
  & {{x}^{2}}-{{y}^{2}}=4.....(1) \\
 & 2xy=3\sqrt{20}.....(2) \\
\end{align}\]
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,
\[\begin{align}
  & \left| {{z}^{2}} \right|=\sqrt{{{\left( 4 \right)}^{2}}+{{\left( 3\sqrt{20i} \right)}^{2}}}=\sqrt{16+180}=\sqrt{196} \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}=\sqrt{196} \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}=14.....(3) \\
\end{align}\]
Solving (1) and (3) we get as follows:
\[\begin{align}
  & {{x}^{2}}-{{y}^{2}}+{{x}^{2}}+{{y}^{2}}=4+14 \\
 & 2{{x}^{2}}=18 \\
 & {{x}^{2}}=\dfrac{18}{2} \\
 & {{x}^{2}}=9 \\
\end{align}\]
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:
\[\begin{align}
  & 2\times 3\times y=3\sqrt{20} \\
 & \Rightarrow 2y=\sqrt{20} \\
 & \Rightarrow y=\dfrac{\sqrt{20}}{2} \\
 & \Rightarrow y=\sqrt{5} \\
\end{align}\]
For x=3, substitute x=-3 in equation (2) we get as follows:
\[\begin{align}
  & 2\times \left( -3 \right)\times y=3\sqrt{20} \\
 & \Rightarrow -2y=\sqrt{20} \\
 & \Rightarrow y=\dfrac{-1}{2}\times \sqrt{20} \\
 & \Rightarrow y=-\sqrt{5} \\
\end{align}\]
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:
\[a\pm bi=\pm \sqrt{\left( {{a}^{2}}-{{b}^{2}} \right)\pm 2abi}\].