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

$\mathit{t}\mathit{a}\mathit{n}\mathit{h}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{\pi i}\mathbf{)}$.

The given expression is $\left(2\pi i\right)$.

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)

$\left(1-\pi i\right)$Use tangent formula.

localid="1658811027663" $\tan h\left(x+yi\right)=\frac{\tan h\left(x\right)+i\tan \left(y\right)}{1+i\tan h\left(x\right)\tan \left(y\right)}$ ….

Use $x=1$and $y=\pi$in the formula (1).

$$\begin{gathered} \tan h\left(1+(-\pi )i\right)=\frac{\tan h\left(1\right)+i\tan \left(-\pi \right)}{1+i\tan h\left(1\right)\tan \left(-\pi \right)} \\ =\tan h\left(1\right) \\ =0.76 \end{gathered}$$

Hence,

The Real part of$\left(1-\pi i\right)$is0.76 .

The Imaginary part of$\left(1-\pi i\right)$is 0.

The absolute value of $\left(1-\pi i\right)$is 0.76.