Q28P - Mathematical Methods in Physical Sciences

Question

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

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

Step-by-Step Solution

Answer

 

The Real part of $\left(1-\pi i\right)$ is $0.76$.

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

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

Step 1: Given Information.

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

Step 2: Definition of Complex Numberand 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)

Step 3: Find the real, imaginary, absolute value of 1 - x i .

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

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

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.