Question

The square root of $1-i$ is $\sqrt{\dfrac{\sqrt{2}+1}{2}}-i\sqrt{\dfrac{\sqrt{2}-1}{2}}$. If this is true enter 1, else enter 0.

Answer 1

Hint: To obtain whether the statement is true or not we will find the square root of the complex number. Firstly we will assume that the square root of the given number is $a-ib$ then we will solve the obtained equation to get the value of $a,b$ and hence we will get our square root. Finally compare whether the obtained square root is equal to the given square root and get the desired answer.

Complete step-by-step solution:
We have to find the square root of:
$1-i$
Let the square root of above value is $a+ib$ so we get the below equation:
$\sqrt{1-i}=\pm \left( a-ib \right)$…..$\left( 1 \right)$
Now we will solve the above equation to get the value of $a,b$
Firstly take square of both side values as follows:
 $\begin{align}
  & {{\left( \sqrt{1-i} \right)}^{2}}={{\left( a-ib \right)}^{2}} \\
 & \Rightarrow {{\left( 1-i \right)}^{\dfrac{1}{2}\times 2}}={{a}^{2}}-2\times a\times ib+{{\left( -ib \right)}^{2}} \\
 & \Rightarrow 1-i={{a}^{2}}-2iab+{{i}^{2}}{{b}^{2}} \\
 & \therefore 1-i={{a}^{2}}-{{b}^{2}}-2iab \\
\end{align}$
Now we will compare the real and imaginary parts as follows:
${{a}^{2}}-{{b}^{2}}=1$…..$\left( 2 \right)$
$2ab=1$…..$\left( 3 \right)$
Next we will find the value of ${{a}^{2}}+{{b}^{2}}$ by using equation (2) and (3) as follows:
${{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}={{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}+4{{a}^{2}}{{b}^{2}}$
$\begin{align}
  & {{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}={{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}+{{\left( 2ab \right)}^{2}} \\
 & \Rightarrow {{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}={{\left( 1 \right)}^{2}}+{{\left( 1 \right)}^{2}} \\
\end{align}$
$\therefore \left( {{a}^{2}}+{{b}^{2}} \right)=\sqrt{2}$….$\left( 4 \right)$
On adding equation (2) and (4) we get,
$\begin{align}
  & 2{{a}^{2}}=1+\sqrt{2} \\
 & \Rightarrow {{a}^{2}}=\dfrac{1+\sqrt{2}}{2} \\
 & \therefore a=\pm \sqrt{\dfrac{1+\sqrt{2}}{2}} \\
\end{align}$
On subtracting equation (2) from (4) we get,
$\begin{align}
  & 2{{b}^{2}}= {\sqrt{2}-1} \\
 & \Rightarrow {{b}^{2}}=\dfrac{\sqrt{2}-1}{2} \\
 & \therefore b=\pm \sqrt{\dfrac{\sqrt{2}-1}{2}} \\
\end{align}$
So we will take the value as:
$\begin{align}
  & a=\sqrt{\dfrac{\sqrt{2}+1}{2}} \\
 & b=\sqrt{\dfrac{\sqrt{2}-1}{2}} \\
\end{align}$
Put above value in equation (1) we get,
$\sqrt{1-i}=\sqrt{\dfrac{\sqrt{2}+1}{2}}-i\sqrt{\dfrac{\sqrt{2}-1}{2}}$
Hence the given statement is true enter 1.

Note: Complex numbers are those that can be expressed in the form of $a+ib$ where $a$ is the real part and $b$ is the imaginary part. Also ${{i}^{2}}=-1$ as $i=\sqrt{-1}$. To find the value of the square root of any complex number we always let it in the form of $a+ib$ the sign in between can be changed depending on the value and then find out the value of $a,b$.