Q17P - Mathematical Methods in Physical Sciences

Question

Show that tan z never takes the values $\mathbf{\pm }\mathit{i}$. Hint: Try to solve the equation tan z  = i and find that it leads to a contradiction.

Step-by-Step Solution

Answer

It has been proved that tan z never takes the values $\pm i$.

Step 1: Given Information.

The given expression is tan z.

Step 2: Meaning of rectangular form.

Represent the complex number in rectangular form means writing the given complex number in the form of x x + iy in which x is the real part and y is the imaginary part.

Step 3: Solve the equation tan z = i .

Solve the given equation tan z = i in the hint.

$$\begin{gathered} \frac{\left[e^{\left(zi\right)}-e^{\left(-zi\right)}\right]}{i\left[e^{\left(zi\right)}+e^{\left(-zi\right)}\right]}=i \\ e^{\left(zi\right)}-e^{\left(-zi\right)}=-e^{\left(zi\right)}-e^{\left(-zi\right)} \\ 2e^{\left(zi\right)}=0 \end{gathered}$$

Previous equation has no exact solution because if we try to take   it gives us infinity which means it's not valid for tan (z) to take $\pm i$ .

 $$\begin{gathered} \frac{\left[e^{\left(zi\right)}-e^{\left(-zi\right)}\right]}{i\left[e^{\left(zi\right)}+e^{\left(-zi\right)}\right]}=i \\ e^{\left(zi\right)}-e^{\left(-zi\right)}=-e^{\left(zi\right)}-e^{\left(-zi\right)} \\ 2e^{\left(-2z\right)}=0 \end{gathered}$$

Previous equation has no exact solution because if we try to take   it gives us infinity which means it's not valid for tan (z) to take $\pm i$.

The final equation is $e^{\left(2z\right)}=0$ which has no exact value.