Question

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

Answer 1

Step 1: Prove that the given statement is true

Let $\sqrt{(1-\mathrm{i})}=\pm (\mathrm{x}-\mathrm{iy})\dots \left(\mathrm{i}\right)$
Square both sides:
$1-\mathrm{i}={\mathrm{x}}^2-{\mathrm{y}}^2-2\mathrm{ixy}$
Compare real and imaginary coefficients:
We get ${\mathrm{x}}^2-{\mathrm{y}}^2=1,2\mathrm{xy}=1...\left(\mathrm{ii}\right)$

And From algebraic identity:
$$\begin{gathered} {\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}^2={\left({\mathrm{x}}^2-{\mathrm{y}}^2\right)}^2+4{\mathrm{x}}^2{\mathrm{y}}^2 \\ \Rightarrow {\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}^2=1^2+1^2 \\ \Rightarrow {\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}^2=2 \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2=\sqrt{2}...\left(\mathrm{iii}\right) \end{gathered}$$

Hence, from $\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iii}\right)$
${\mathrm{x}}^2+{\mathrm{y}}^2=\sqrt{2}$ and $x^2-y^2=1$
Add $x^2-y^2=1$ and $x^2-y^2=1$ :

$\begin{array}{rl}& 2{\mathrm{x}}^2=\sqrt{2}+1\\ \Rightarrow & {\mathrm{x}}^2=\frac{(\sqrt{2}+1)}{2}\\ \Rightarrow & \mathrm{x}=\sqrt{\frac{(\sqrt{2}+1)}{2}}...\left(\mathrm{iv}\right)\end{array}$

Put$x^2=\frac{(\sqrt{2}+1)}{2}$ in $x^2-y^2=1$:

$$\begin{gathered} \mathit{}{x}^{\mathit{2}}\mathit{-}{y}^{\mathit{2}}\mathit{=}1 \\ \Rightarrow y^2={\mathrm{x}}^2-1 \\ \Rightarrow y^2=\frac{(\sqrt{2}+1)}{2}-1 \\ \Rightarrow y^{\mathit{2}}=\frac{(\sqrt{2}-1)}{2} \\ \Rightarrow y=\sqrt{\frac{(\sqrt{2}-1)}{2}}...\left(\mathrm{v}\right) \end{gathered}$$
Step 2: Compute the required value.

Put value $x=\sqrt{\frac{(\sqrt{2}+1)}{2}}\mathrm{and}y=\sqrt{\frac{(\sqrt{2}-1)}{2}}$ from $\left(\mathrm{iv}\right)\mathrm{and}\left(\mathrm{v}\right)$in $\sqrt{(1-\mathrm{i})}=\pm (\mathrm{x}-\mathrm{iy})$

$$\begin{gathered} \sqrt{(1-\mathrm{i})}=\pm (\mathrm{x}-\mathrm{iy}) \\ \Rightarrow \sqrt{(1-i)}=\sqrt{\frac{(\sqrt{2}+1)}{2}}-i\sqrt{\frac{(\sqrt{2}-1)}{2}} \end{gathered}$$

Thus ,the given statement is true and hence enter $1.$