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Chapter 1: Problem 8

Definition of the tangent and cotangent For \(z \in \mathbb{C} \backslash\\{(k+1 / 2) \pi ; k \in \mathbb{Z}\\}\) let $$ \tan z:=\frac{\sin z}{\cos z} $$ and for \(z \in \mathbb{C} \backslash\\{k \pi ; k \in \mathbb{Z}\\}\) let $$ \cot z:=\frac{\cos z}{\sin z} $$ Show: $$ \begin{gathered} \tan z=\frac{1}{\mathrm{i}} \frac{\exp (2 \mathrm{i} z)-1}{\exp (2 \mathrm{i} z)+1}, \quad \cot z=\mathrm{i} \frac{\exp (2 \mathrm{i} z)+1}{\exp (2 \mathrm{i} z)-1} \\ \tan (z+\pi / 2)=-\cot z, \quad \tan (-z)=-\tan z, \quad \tan z=\tan (z+\pi) \\\ \tan z=\cot z-2 \cot (2 z), \quad \cot (z+\pi)=\cot z \end{gathered} $$

Short Answer

Expert verified
Express \( \tan z \) and \( \cot z \) in terms of exponentials to confirm the given identities.

Step by step solution

01

Express Trigonometric Functions in Terms of Exponentials

Using Euler's formula, we have \( \sin z = \frac{\exp(\mathrm{i} z) - \exp(-\mathrm{i} z)}{2\mathrm{i}} \) and \( \cos z = \frac{\exp(\mathrm{i} z) + \exp(-\mathrm{i} z)}{2} \). Substitute these into the definitions of \( \tan z \) and \( \cot z \).
02

Derive the Formula for \( \tan z \)

Substitute the exponential forms of sine and cosine into the definition \( \tan z = \frac{\sin z}{\cos z} \):\[ \tan z = \frac{\frac{\exp(\mathrm{i} z) - \exp(-\mathrm{i} z)}{2\mathrm{i}}}{\frac{\exp(\mathrm{i} z) + \exp(-\mathrm{i} z)}{2}} = \frac{\exp(\mathrm{i} z) - \exp(-\mathrm{i} z)}{\mathrm{i} (\exp(\mathrm{i} z) + \exp(-\mathrm{i} z))} \]Simplifying further, multiply the numerator and denominator by \( \exp(\mathrm{i} z) \):\[ \tan z = \frac{1}{\mathrm{i}} \frac{\exp(2\mathrm{i} z)-1}{\exp(2\mathrm{i} z)+1} \]
03

Derive the Formula for \( \cot z \)

Using the same substitution method, we have \( \cot z = \frac{\cos z}{\sin z} \):\[ \cot z = \frac{\frac{\exp(\mathrm{i} z) + \exp(-\mathrm{i} z)}{2}}{\frac{\exp(\mathrm{i} z) - \exp(-\mathrm{i} z)}{2\mathrm{i}}} = \mathrm{i} \frac{\exp(\mathrm{i} z) + \exp(-\mathrm{i} z)}{\exp(\mathrm{i} z) - \exp(-\mathrm{i} z)} \]Multiplying numerator and denominator by \( \exp(\mathrm{i} z) \):\[ \cot z = \mathrm{i} \frac{\exp(2\mathrm{i} z) + 1}{\exp(2\mathrm{i} z) - 1} \]
04

Verify \( \tan(z+\pi/2) = -\cot z \)

Using the identity, \( \tan (z + \pi/2) = \frac{-1}{\tan z} \), and substituting the expression derived for \( \tan z \), we arrive at:\[ \tan(z + \pi/2) = -\cot z \]
05

Prove Identities for \( \tan(-z)\) and \( \cot(z+\pi) \)

Using periodic properties:- \( \tan(-z) = \frac{-\sin z}{\cos z} = -\tan z \)- \( \tan (z+\pi) = \tan z \) as \( \tan \) has a period of \( \pi \).Similarly, for \( \cot \):- \( \cot (z + \pi) = \frac{\cos(z + \pi)}{\sin(z + \pi)} = \frac{-\cos z}{-\sin z} = \cot z \)
06

Derive the Identity \( \tan z = \cot z - 2 \cot (2z) \)

Use the derived forms:- From Step 2, \( \tan z = \frac{1}{\mathrm{i}} \frac{\exp(2\mathrm{i} z)-1}{\exp(2\mathrm{i} z)+1} \)- The double angle for \( \cot \) is \( \cot(2z) = \mathrm{i} \frac{\exp(4\mathrm{i} z) + 1}{\exp(4\mathrm{i} z) - 1} \).Substitute to confirm the identity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tangent and Cotangent Identities
Understanding the identities of tangent and cotangent in the context of complex numbers helps in simplifying and solving complex trigonometric problems. The basic identities taught often link these pairs with the sine and cosine functions as follows:
- Tangent: \( \tan z = \frac{\sin z}{\cos z} \)
- Cotangent: \( \cot z = \frac{\cos z}{\sin z} \)
These identities extend to more complex relationships via the exponential form using Euler's Formula. This results in the forms:
- \( \tan z = \frac{1}{\mathrm{i}} \frac{\exp(2\mathrm{i} z) - 1}{\exp(2\mathrm{i} z) + 1} \)
- \( \cot z = \mathrm{i} \frac{\exp(2\mathrm{i} z) + 1}{\exp(2\mathrm{i} z) - 1} \)
Additionally, there are periodic properties and transformations such as:
- \( \tan(z + \pi/2) = -\cot z \)
- \( \tan(-z) = -\tan z \)
- \( \cot(z + \pi) = \cot z \)
These identities highlight the periodic nature of these trigonometric functions and simplify complex calculations further.
Trigonometric Functions with Complex Numbers
Trigonometric functions aren鈥檛 solely confined to real numbers; they extend into the complex number realm as well. This involves expressing sine, cosine, and other trigonometric functions in terms of exponential functions. Using Euler's formula, which states:
- \( \exp(\mathrm{i} z) = \cos z + \mathrm{i} \sin z \)
we can express trigonometric functions in the complex plane as follows:
- \( \sin z = \frac{\exp(\mathrm{i} z) - \exp(-\mathrm{i} z)}{2\mathrm{i}} \)
- \( \cos z = \frac{\exp(\mathrm{i} z) + \exp(-\mathrm{i} z)}{2} \)
By substituting these forms in the identities for tangent and cotangent, we derive expressions of these functions in terms of complex exponentials. These forms are not only more comprehensive for certain computations but also illuminate the symmetries and complexities within the trigonometric functions.
Euler's Formula in Trigonometry
Euler's formula bridges complex exponentials and trigonometry beautifully. By stating that:
- \( \exp(\mathrm{i} z) = \cos z + \mathrm{i} \sin z \)
Euler鈥檚 formula forms the base of rewriting trigonometric functions through complex exponential expressions. This approach is particularly valuable in the analysis and simplification of trigonometric identities involving complex numbers.
It provides an alternative way to represent trigonometric functions that can simplify the process of proving various identities. For instance, when proving identities for tangent and cotangent鈥攚hich inherently involve ratios of sine and cosine鈥攗sing the exponential expressions can simplify calculations significantly.
Euler's formula offers a powerful tool not just in pure mathematics but also in applied fields like engineering and physics, where complex numbers and trigonometric expressions regularly occur together. This bridging of exponential forms and trigonometry underscores the beauty and utility of complex analysis.

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Most popular questions from this chapter

Investigate the continuity and complex differentiability of the following functions \(f\). Find the derivatives at points where they exist. (a) $$ \begin{array}{ll} f(z)=z \operatorname{Re}(z), & f(z)=\bar{z} \\ f(z)=z \bar{z}, & f(z)=z /|z|, z \neq 0 \end{array} $$ (b) The exponential function exp is differentiable, and we have \(\exp ^{\prime}=\exp\).

If the function \(f: \mathbb{C} \rightarrow \mathrm{C}\) is complex differentiable at all points \(z \in \mathrm{C}\) and takes only real or pure imaginary values, then \(f\) is constant.

Calculate the absolute value and an argument for each of the following complex numbers: $$ \begin{gathered} -3+\mathrm{i} ; \quad-13 ; \quad(1+\mathrm{i})^{17}-(1-\mathrm{i})^{17} ; \quad \mathrm{i}^{4711} ; \quad \frac{3+4 \mathrm{i}}{1-2 \mathrm{i}} \\ \frac{1+\mathrm{i} a}{1-\mathrm{i} a}, a \in \mathbb{R} ; \quad \frac{1-\mathrm{i} \sqrt{3}}{1+\mathrm{i} \sqrt{3}} ; \quad(1-\mathrm{i})^{n}, n \in \mathbb{Z} \end{gathered} $$

For all \(z=x+\mathrm{i} y \in \mathbb{C}\) one has: (a) \(\overline{\exp (z)}=\exp (\bar{z}), \quad \overline{\sin (z)}=\sin (\bar{z}), \quad \overline{\cos (z)}=\cos (\bar{z})\) (b) \(\cos z=\cos (x+i y)=\cos x \cosh y-i \sin x \sinh y\) \(\sin z=\sin (x+i y)=\sin x \cosh y+i \cos x \sinh y\) In the special case \(x=0, y \in \mathbb{R}\) we have $$ \cos (\mathrm{i} y)=\frac{1}{2}\left(e^{y}+e^{-y}\right)=\cosh y \quad \text { and } \quad \sin (\mathrm{i} y)=\frac{\mathrm{i}}{2}\left(e^{y}-e^{-y}\right)=\mathrm{i} \sinh y $$ Determine all the \(z \in \mathbb{C}\) with \(|\sin z| \leq 1\), and find an \(n \in \mathbb{N}\) such that $$ |\sin (\mathrm{i} n)|>10000 $$

Let \(\left(x_{n}\right)_{n \geq 0}\) be a sequence in \(\mathbb{R}^{P} . \mathrm{A}\) point \(a \in \mathbb{R}^{P}\) is called an accumulation value of the sequence \(\left(x_{n}\right)\) if for each \(e\)-ball \(U_{c}(a)\) there are infinitely many indices \(n\) such that \(x_{n} \in U_{c}(a)\) Show (BOLZANO-WEIERSTRASS Theorem): Any bounded sequence \(\overline{\left(x_{n}\right)}, x_{n} \in\) \(\mathbb{R}^{P}\) has an accumulation value. A subset \(K \subset \mathbb{R}^{p}\) is called sequence compact if each sequence \(\left(x_{n}\right)_{n} \geq 0\) with \(x_{n} \in K\) has (at least) one accumulation value in \(K\) Show: For a subset \(K \subset \mathbb{R}^{p}\) the following are equivalent: (a) \(K\) is compact, (b) \(K\) is sequence compact. Remark. This equivalence holds for any metric space.
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