Question

If $\cos (u+iv)=x+iy$, then $x^2+y^2+1=?$

• A. ${\cos }^{2}u+\sin h^{2}v$
• B. ${\sin }^{2}u+\cos h^{2}v$
• C. ${\cos }^{2}u+\cos h^{2}v$
• D. ${\sin }^{2}u+\sin h^{2}v$

Answer 1

The correct option is C

${\cos }^{2}u+\cos h^{2}v$

Explanation for the correct option:

Step 1. Find the value of $x^2+y^2+1$:

Given, $\cos (u+iv)=x+iy$

$\Rightarrow$$\cos u\cos \left(iv\right)–\sin u\sin \left(iv\right)=x+iy$

$\Rightarrow$$\cos u\cos hv–i\sin u\sin hv=x+iy$

Compare real and imaginary parts, we get

$x=\cos u\cos hv$ …(1)

$y=–\sin u\sin hv$ …(2)

Step 2. Square and add both side of equation (1) and (2), we get

$\begin{array}{rcl}{x}^2+y^2& =& {\cos }^{2}u\cos h^{2}v+{\sin }^{2}u\sin h^{2}v\\ & =& (1–{\sin }^{2}u)\cos h^{2}v+{\sin }^{2}u(\cos h^{2}v–1)\\ & =& \cos h^{2}v–{\sin }^{2}u\end{array}$ ; $\left[\mathbf{\because }\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\right]$

$\Rightarrow$$\begin{array}{rcl}{x}^2+y^2+1& =& \cos h^{2}v+1–{\sin }^{2}u\\ & =& \mathit{c}\mathit{o}\mathit{s}\mathbf{}{\mathit{h}}^{\mathbf{2}}\mathbf{}\mathit{v}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{}\mathit{u}\end{array}$

Hence, Option ‘C’ is Correct.