Question

Question: Find the real part, the imaginary part, and the absolute value of

$\mathbf{sin}\mathbf{(}\mathit{x}\mathbf{-}\mathit{y}\mathbf{i}\mathbf{)}$.

Short answer

The Real part of$\sin (x-y\mathrm{i})$is $\sin \left(x\right)\cosh \left(y\right)$.

The Imaginary part of $\sin (x-y\mathrm{i})$is $-\cos \left(x\right)\sinh \left(y\right)$.

The absolute value of$\sin (x-y\mathrm{i})$ is $\sqrt{{\sin }^2\left(x\right){\sinh }^2\left(y\right)}$.

Step-by-step solution

Given Information.

The given expression is $\sin (\mathrm{x}-\mathrm{yi})$.

Definition of Complex Number and Hyperbolic functions.

The domain of a function refers to the range of values that can be plugged into it. This is the set of x values in a function like f(x). The range of a function is the set of values that the function can take. This is the set of values that the function produces when we enter x value . The y values are what you're looking for.

The basic hyperbolic functions are:

1. Hyperbolic sine(sinh)
2. Hyperbolic cosine(cosh)
3. Hyperbolic tangent(tanh)

Find the real , imaginary , absolute value of sin(x-yi).

Let u = sin(x - yi).

Use sine formula.

$\sinh (\mathrm{a}+\mathrm{bi})=\sin \left(\mathrm{a}\right)\cosh \left(\mathrm{b}\right)+\mathrm{icos}\left(\mathrm{a}\right)+\sinh \left(\mathrm{b}\right)$ ….

Use a = x and

b = -y In the formula .

$$\begin{gathered} \mathrm{Re}\left(\mathrm{u}\right)=\sin \left(\mathrm{a}\right)\cosh \left(\mathrm{b}\right) \\ =\sin \left(\mathrm{x}\right)\cosh \left(-\mathrm{y}\right) \\ =\sin \left(\mathrm{x}\right)\cosh \left(\mathrm{y}\right) \\ \mathrm{Im}\left(\mathrm{u}\right)=\cos \left(\mathrm{a}\right)\sinh \left(\mathrm{b}\right) \\ =\cos \left(\mathrm{x}\right)\sinh (-\mathrm{y}) \\ =-\cos \left(\mathrm{x}\right)\sinh \left(\mathrm{y}\right) \\ \left|\mathrm{u}\right|=\sqrt{{\mathrm{Re}}^2\left(\mathrm{u}\right)+{\mathrm{Im}}^2\left(\mathrm{u}\right)} \\ =\sqrt{{\left(\sin \left(\mathrm{x}\right)\cosh \left(\mathrm{y}\right)\right)}^2+{\left(-\cos \left(\mathrm{x}\right)\sinh \left(\mathrm{y}\right)\right)}^2} \end{gathered}$$

Let …….

$$\begin{gathered} v={\left(\sin \left(\mathrm{x}\right)\cosh \left(\mathrm{y}\right)\right)}^2+{\left(-\cos \left(\mathrm{x}\right)\sinh \left(\mathrm{y}\right)\right)}^2 \\ {\cosh }^2\left(\mathrm{y}\right)=1+{\sinh }^2\left(\mathrm{y}\right) \\ {\cos }^2\left(\mathrm{y}\right)=1+{\sin }^2\left(\mathrm{y}\right) \end{gathered}$$

Start the simplification of equation (2).

$$\begin{gathered} \mathrm{v}={\left(\sin \left(\mathrm{x}\right)\cosh \left(\mathrm{y}\right)\right)}^2+{\left(-\cos \left(\mathrm{x}\right)\sinh \left(\mathrm{y}\right)\right)}^2 \\ ={\sin }^2\left(\mathrm{x}\right)\left[1+{\sinh }^2\left(\mathrm{y}\right)\right]+\left[1-{\sin }^2\left(\mathrm{x}\right)\right]{\sinh }^2\left(\mathrm{y}\right) \\ ={\sin }^2\left(\mathrm{x}\right)+{\sin }^2\left(\mathrm{x}\right){\sinh }^2\left(\mathrm{y}\right)+{\sinh }^2\left(\mathrm{y}\right)-{\sin }^2\left(\mathrm{x}\right){\sinh }^2\left(\mathrm{y}\right) \\ ={\sin }^2\left(\mathrm{x}\right)+{\sinh }^2\left(\mathrm{y}\right) \end{gathered}$$

Put the value of v in (1).

role="math" localid="1658811806900" $$\begin{gathered} \left|\mathrm{u}\right|=\sqrt{{\mathrm{Re}}^2\left(\mathrm{u}\right)+{\mathrm{Im}}^2\left(\mathrm{u}\right)} \\ =\sqrt{{\left(\sin \left(\mathrm{x}\right)\cosh \left(\mathrm{y}\right)\right)}^2+{\left(-\cos \left(\mathrm{x}\right)\sinh \left(\mathrm{y}\right)\right)}^2} \\ =\sqrt{{\sin }^2\left(\mathrm{x}\right)+{\sinh }^2\left(\mathrm{y}\right)} \end{gathered}$$

Hence,

The Real part of (x - yi) is sin(x)cosh(y).

The Imaginary part of (x - yi) is $-\cos \left(\mathrm{x}\right)\sinh \left(\mathrm{y}\right)$.

The absolute value of (x - yi) is role="math" localid="1658811799720" $\sqrt{{\sin }^2\left(\mathrm{x}\right)+{\sinh }^2\left(\mathrm{y}\right)}$.