Find the square root of the given complex number:
$z=-16+30i$
Answer
677.1k+ views
Hint: Here, we will assume that the square root of the given complex number to be x+iy. It means that the given complex number is the square of x+iy. Using this we can calculate the square root of -16 + 30i.
Complete step-by-step answer:
A complex number is a number that can be expressed in the form a+ib, where a and b are real numbers and I is the solution of the equation ${{x}^{2}}=-1$. Since, no real number satisfies this equation, $i$ is called an imaginary number.
Let us assume that the square root of the given complex number that is $z=-16+30i$ is $x+iy$.
Since, $z=-16+30i$
We can write:
On comparing the real part and imaginary part on both sides of the above equation, we get:
$\begin{align}
& {{x}^{2}}-{{y}^{2}}=-16........\left( 1 \right) \\
& 2xy=30.............\left( 2 \right) \\
\end{align}$
Now, from equation (1), we can write:
$\begin{align}
& \left( {{x}^{2}}-{{y}^{2}} \right)={{\left( -16 \right)}^{2}} \\
& \Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-4{{x}^{2}}{{y}^{2}}=256 \\
\end{align}$
Using equation (2), that is putting the value of 2xy from equation (2) in the above equation, we get:
$\begin{align}
& {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-{{\left( 30 \right)}^{2}}=256 \\
& \Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-900=256 \\
& \Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=1156 \\
& \Rightarrow \left( {{x}^{2}}+{{y}^{2}} \right)=\sqrt{1156}=34.............\left( 3 \right) \\
\end{align}$
On adding equation (1) and (3), we get:
$\begin{align}
& {{x}^{2}}-{{y}^{2}}+{{x}^{2}}+{{y}^{2}}=-16+34 \\
& \Rightarrow 2{{x}^{2}}=18 \\
& \Rightarrow {{x}^{2}}=9 \\
& \Rightarrow x=\pm 3 \\
\end{align}$
When x=3, on putting x = 3 in equation (2), we get:
$\begin{align}
& 2\times 3\times y=30 \\
& \Rightarrow y=\dfrac{30}{6}=5 \\
\end{align}$
On putting x = -3, in equation (2), we get:
$\begin{align}
& 2\times \left( -3 \right)\times y=30 \\
& \Rightarrow y=\dfrac{-30}{6}=-5 \\
\end{align}$
So, the values of x are 3, -3 and that of y are 5 and -5.
Hence, the square root of the complex number $-16+30i$ are $3+5i$ and $-3-5i$.
Note: Students should note here that the value of ${{i}^{2}}$ is -1. Thus, while calculating the value of ${{\left( x+iy \right)}^{2}}$, it should be kept in mind that ${{i}^{2}}$ should be replaced by -1. The calculations must be done properly to avoid mistakes.
Complete step-by-step answer:
A complex number is a number that can be expressed in the form a+ib, where a and b are real numbers and I is the solution of the equation ${{x}^{2}}=-1$. Since, no real number satisfies this equation, $i$ is called an imaginary number.
Let us assume that the square root of the given complex number that is $z=-16+30i$ is $x+iy$.
Since, $z=-16+30i$
We can write:
On comparing the real part and imaginary part on both sides of the above equation, we get:
$\begin{align}
& {{x}^{2}}-{{y}^{2}}=-16........\left( 1 \right) \\
& 2xy=30.............\left( 2 \right) \\
\end{align}$
Now, from equation (1), we can write:
$\begin{align}
& \left( {{x}^{2}}-{{y}^{2}} \right)={{\left( -16 \right)}^{2}} \\
& \Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-4{{x}^{2}}{{y}^{2}}=256 \\
\end{align}$
Using equation (2), that is putting the value of 2xy from equation (2) in the above equation, we get:
$\begin{align}
& {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-{{\left( 30 \right)}^{2}}=256 \\
& \Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-900=256 \\
& \Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=1156 \\
& \Rightarrow \left( {{x}^{2}}+{{y}^{2}} \right)=\sqrt{1156}=34.............\left( 3 \right) \\
\end{align}$
On adding equation (1) and (3), we get:
$\begin{align}
& {{x}^{2}}-{{y}^{2}}+{{x}^{2}}+{{y}^{2}}=-16+34 \\
& \Rightarrow 2{{x}^{2}}=18 \\
& \Rightarrow {{x}^{2}}=9 \\
& \Rightarrow x=\pm 3 \\
\end{align}$
When x=3, on putting x = 3 in equation (2), we get:
$\begin{align}
& 2\times 3\times y=30 \\
& \Rightarrow y=\dfrac{30}{6}=5 \\
\end{align}$
On putting x = -3, in equation (2), we get:
$\begin{align}
& 2\times \left( -3 \right)\times y=30 \\
& \Rightarrow y=\dfrac{-30}{6}=-5 \\
\end{align}$
So, the values of x are 3, -3 and that of y are 5 and -5.
Hence, the square root of the complex number $-16+30i$ are $3+5i$ and $-3-5i$.
Note: Students should note here that the value of ${{i}^{2}}$ is -1. Thus, while calculating the value of ${{\left( x+iy \right)}^{2}}$, it should be kept in mind that ${{i}^{2}}$ should be replaced by -1. The calculations must be done properly to avoid mistakes.
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