Question

Multiply ${{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}}$ by ${{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}}$.

Answer 1

Hint:Here, $\left( {{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}} \right)\left( {{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}} \right)$ is of the form $(a+b)(a-b)$ which is ${{a}^{2}}-{{b}^{2}}$ and we have to substitute for $\sqrt{-1}=i$, and then apply the formula ${{e}^{i\theta }}=\cos \theta +i\sin \theta$ and hence, do the simplification.

Complete Step-by-step answer:
Here, we have to calculate $\left( {{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}} \right)\left( {{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}} \right)$.
That is, it is of the form $(a+b)(a-b)$. We know the formula that:
$(a+b)(a-b)={{a}^{2}}-{{b}^{2}}$
Here, we have $a={{e}^{\sqrt{-1}}}$ and $b={{e}^{-\sqrt{-1}}}$.
Therefore, by applying the formula we get the equation:
$\left( {{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}} \right)\left( {{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}} \right)={{\left( {{e}^{\sqrt{-1}}} \right)}^{2}}-{{\left( {{e}^{-\sqrt{-1}}} \right)}^{2}}$ …. (1)
We know that $\sqrt{-1}$ is a complex number and it is considered as the imaginary value $i$. Hence, we can write:
$\sqrt{-1}=i$
Therefore, our equation (1) becomes:
$\left( {{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}} \right)\left( {{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}} \right)={{\left( {{e}^{i}} \right)}^{2}}-{{\left( {{e}^{-i}} \right)}^{2}}$ …. (2)
We also know the property that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$.
 Therefore, we can say that ${{\left( {{e}^{i}} \right)}^{2}}={{e}^{2i}}$ and ${{\left( {{e}^{-i}} \right)}^{2}}={{e}^{-2i}}$.
Hence, our equation (2) becomes:
$\left( {{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}} \right)\left( {{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}} \right)={{e}^{2i}}-{{e}^{-2i}}$ …. (3)
The RHS in the above equation is of the form ${{e}^{i\theta }}$.
We know the formula that:
${{e}^{i\theta }}=\cos \theta +i\sin \theta$
 We have ${{e}^{2i}}={{e}^{i2}}$
By applying this formula we will get:
${{e}^{2i}}=\cos 2+i\sin 2$ …… (4)
Similarly, we can say that,
$\begin{align}
  & {{e}^{-2i}}={{e}^{i(-2)}} \\
 & {{e}^{-2i}}=\cos (-2)+i\sin (-2) \\
\end{align}$
We know that $\cos (-x)=\cos x$ and $\sin (-x)=-\sin x$. Hence, we will get:
$\cos (-2)=\cos 2$ and $\sin (-2)=-\sin 2$
Therefore, our equation becomes:
${{e}^{-2i}}=\cos 2-i\sin 2$ ….. (5)
Hence, by substituting equation (4) and equation (5) in equation (3) we obtain the equation:
$\left( {{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}} \right)\left( {{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}} \right)=\cos 2+i\sin 2-(\cos 2-i\sin 2)$
Now, take $\cos 2-i\sin 2$outside the bracket we get:
$\left( {{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}} \right)\left( {{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}} \right)=\cos 2+i\sin 2-\cos 2+i\sin 2$
 In the next step, we have to cancel $\cos 2$ and $i\sin 2+i\sin 2=2i\sin 2$, so we obtain the equation: $\left( {{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}} \right)\left( {{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}} \right)=2i\sin 2$
Therefore, by multiplying ${{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}}$ by ${{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}}$ we obtain the solution, $2i\sin 2$, where $i$ is the imaginary number whose value is taken as, $i=\sqrt{-1}$.

Note: Here, after getting $\left( {{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}} \right)\left( {{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}} \right)={{e}^{2i}}-{{e}^{-2i}}$, you should not stop. Next, you change this equation into the form ${{e}^{i\theta }}=\cos \theta +i\sin \theta $ and reduce it into a much simpler form. You also should have an idea regarding the properties of exponential functions.