Question

Find the square root of (- 20 – 20i).

Answer 1

Hint: Take \[\sqrt{\left( -21-20i \right)}=a+bi\]. Squaring and simplifying, find the real and imaginary part. Thus find the value of a and b by solving the equation formed. Substitute the values in (a + bi) to get the square root.
Complete step by step answer:
Every complex number has a complex square root. We have been asked to find the square root of the complex number (-21 – 20i). We know that all square roots of the number will satisfy the equation \[-21-20i={{x}^{2}}\] by definition of a square root.
We also know that x can be expressed as (a + bi), where a and b are real numbers. Since, the square roots of a complex number are always complex.
So let us take \[\sqrt{\left( -21-20i \right)}=\left( a+bi \right)\]
Let us square on both sides.
\[-21-20i={{\left( a+bi \right)}^{2}}\]
We know that, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]
\[\Rightarrow -21-20i={{a}^{2}}+2abi+{{\left( ib \right)}^{2}}\] (\[\because {{i}^{2}}=-1\])
\[\Rightarrow -21-20i={{a}^{2}}-{{b}^{2}}+2abi\]
Now both sides of the equation are the same. Let us compare the real part of the equation. Thus we get,
\[{{a}^{2}}-{{b}^{2}}=-21-(1)\]
Now let us compare the imaginary part of the equation. We get,
\[\begin{align}
  & 2abi=-20i\Rightarrow 2ab=-20 \\
 & \Rightarrow ab=-10-(2) \\
\end{align}\]
We now have two equations with 2 unknowns. We can solve these simultaneous equations for a and b.
From (2) we can say that, \[b=\dfrac{-10}{a}\].
Now let us substitute this in equation (1).
\[\begin{align}
  & {{a}^{2}}-{{b}^{2}}=-21 \\
 & \Rightarrow {{a}^{2}}-{{\left( \dfrac{-10}{a} \right)}^{2}}=-21 \\
 & \Rightarrow {{a}^{2}}-\dfrac{100}{{{a}^{2}}}=-21 \\
 & \Rightarrow {{a}^{4}}-100=-21{{a}^{2}} \\
 & \Rightarrow {{a}^{4}}+21{{a}^{2}}-100=0 \\
\end{align}\]
We can split the second term \[21{{a}^{2}}\] as \[\left( 25-4 \right){{a}^{2}}\].
\[\begin{align}
  & \therefore {{a}^{2}}+\left( 25-4 \right){{a}^{2}}-100=0 \\
 & {{a}^{2}}+25{{a}^{2}}-4{{a}^{2}}-100=0 \\
 & {{a}^{2}}\left( {{a}^{2}}+25 \right)-4\left( {{a}^{2}}+25 \right)=0 \\
 & \Rightarrow \left( {{a}^{2}}-4 \right)\left( {{a}^{2}}+25 \right)=0 \\
\end{align}\]
Thus \[{{a}^{2}}-4=0\]
\[\begin{align}
  & \Rightarrow {{a}^{2}}=4 \\
 & \therefore a=\sqrt{4}=\pm 2 \\
\end{align}\]
And \[{{a}^{2}}+25=0\]
\[{{a}^{2}}=-25\]
We assumed a to be a real number. So, \[{{a}^{2}}=-25\] has no solutions.
Thus we got a as 2 and -2.
\[\therefore \] a = 2 and a = -2.
Now let us substitute the value of a in equation (2).
\[\begin{align}
  & ab=-10 \\
 & \Rightarrow b=\dfrac{-10}{a} \\
\end{align}\]
When a = 2, \[b=\dfrac{-10}{2}=-5\]
            a = -2, \[b=\dfrac{-10}{\left( -2 \right)}=5\]
Thus by putting a and b in (a + bi), we get the square roots of (-21 - 20i) as (2 - 5i) and (-2 + 5i) \[=\pm \left( 2-5i \right)\].
\[\therefore \] The square root is (-21 – 20i) = (2 – 5i) and (-2 + 5i).

Note: We can also solve it as,
We got real part as, \[{{a}^{2}}-{{b}^{2}}=-21-(1)\]
Imaginary part as, \[2xy=-20-(2)\]
\[\begin{align}
  & {{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}={{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}+4{{a}^{2}}{{b}^{2}} \\
 & {{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}={{\left( -21 \right)}^{2}}+{{\left( -20 \right)}^{2}}=841 \\
 & \therefore {{a}^{2}}+{{b}^{2}}=\sqrt{841}=29-(3) \\
\end{align}\]
Adding (1) and (3), we get
\[\begin{align}
  & 2{{a}^{2}}=8 \\
 & \therefore a=\pm 2 \\
\end{align}\]
Thus we get, \[b=\pm 5\].
\[\therefore \] Square root of -21 – 20i = \[\pm \left( 2-5i \right)\].