Question

If $\cosh \left( u+iv \right)=x+iy$ then $\dfrac{{{x}^{2}}}{{{\cos }^{2}}v}-\dfrac{{{y}^{2}}}{{{\sin }^{2}}v}$ is
A. $0$
B. $-1$
C. $1$
D. $2$

Answer 1

Hint: We will use the formula $\cosh \left( a+ib \right)=\cos \left( ia-b \right)$ to expand the term $\cosh \left( u+iv \right)$ then we use some basic properties of complex sine and cosine functions and convert the expression in the form of $a+ib$ and then we will equate the R.H.S and L.H.S to get the values of $x,y$. By using the values of $x,y$ we will find the value of $\dfrac{{{x}^{2}}}{{{\cos }^{2}}v}-\dfrac{{{y}^{2}}}{{{\sin }^{2}}v}$

Complete step by step answer:
Given that $\cosh \left( u+iv \right)=x+iy$
We know the formula $\cosh \left( a+ib \right)=\cos \left( ia-b \right)$, so
$x+iy=\cos \left( iv-u \right)$
Use the formula $\cos \left( ia-b \right)=\cos ia.\cos b+\sin ia.\sin b$ in the above equation, we have
$x+iy=\cos iu.\cos v+\sin iu.\sin v$
We have basic properties of complex $\sin $ and $\cos $function as $\cos ia=\cosh a$ and $\sin ia=i\sinh a$then the above equation is modified as
 $\begin{align}
  & x+iy=\cosh u.\cos v+i\sinh u.\sin v \\
 & =\left( \cosh u.\cos v \right)+i\left( \sinh u.\sin v \right)
\end{align}$
Equating on both sides we get the values of $x,y$as
$x=\cosh u.\cos v$ and $y=\sinh u.\sin v$
Now the value of $\dfrac{{{x}^{2}}}{{{\cos }^{2}}v}-\dfrac{{{y}^{2}}}{{{\sin }^{2}}v}$ is
$\begin{align}
  & \dfrac{{{x}^{2}}}{{{\cos }^{2}}v}-\dfrac{{{y}^{2}}}{{{\sin }^{2}}v}=\dfrac{{{\left( \cosh u.\cos v \right)}^{2}}}{{{\cos }^{2}}v}-\dfrac{{{\left( \sinh u.\sin v \right)}^{2}}}{{{\sin }^{2}}v} \\
 & ={{\cosh }^{2}}v-{{\sinh }^{2}}v
\end{align}$
We know $\cosh x=\left( \dfrac{{{e}^{x}}+{{e}^{-x}}}{2} \right)$ and $\sinh x=\left( \dfrac{{{e}^{x}}-{{e}^{-x}}}{2} \right)$so the above equation modified as
\[\begin{align}
  & \dfrac{{{x}^{2}}}{{{\cos }^{2}}v}-\dfrac{{{y}^{2}}}{{{\sin }^{2}}v}={{\cosh }^{2}}v-{{\sinh }^{2}}v \\
 & ={{\left( \dfrac{{{e}^{x}}+{{e}^{-x}}}{2} \right)}^{2}}-{{\left( \dfrac{{{e}^{x}}-{{e}^{-x}}}{2} \right)}^{2}} \\
 & =\dfrac{{{\left( {{e}^{x}}+{{e}^{-x}} \right)}^{2}}-{{\left( {{e}^{x}}-{{e}^{-x}} \right)}^{2}}}{4} \\
 & =\dfrac{{{e}^{2x}}+{{e}^{-2x}}+2.{{e}^{x-x}}-\left( {{e}^{2x}}+{{e}^{-2x}}-2.{{e}^{x-x}} \right)}{4} \\
 & =\dfrac{{{e}^{2x}}+{{e}^{-2x}}+2-{{e}^{2x}}-{{e}^{-2x}}+2}{4} \\
 & =1
\end{align}\]

So, the correct answer is “Option C”.

Note: The hyperbolic equations look similar so we have to be careful while we are using the equations. You can directly use the formula ${{\cosh }^{2}}x-{{\sinh }^{2}}x=1$ instead of using the formulas $\cosh x=\left( \dfrac{{{e}^{x}}+{{e}^{-x}}}{2} \right)$, $\sinh x=\left( \dfrac{{{e}^{x}}-{{e}^{-x}}}{2} \right)$