Complex Numbers

Hyperbolic functions of a complex variable

Hyperbolic functions of a complex variable connect exponential definitions with circular trigonometry through Euler's relations. Their properties include periodicity, real and imaginary parts, and inverse relations.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Definition and Fundamental Identity Principle for Complex Hyperbolic Functions

For any complex variable $z = x + iy$, the complex hyperbolic sine and cosine are defined via exponential series as \begin{aligned} \sinh(z) &= \frac{e^z - e^{-z}}{2}, \\ \cosh(z) &= \frac{e^z + e^{-z}}{2} \end{aligned} satisfying the fundamental fundamental identity $\cosh^2(z) - \sinh^2(z) = 1$.

📐 Geometric Interpretation

While circular functions parameterize the unit circle $x^2 + y^2 = 1$, complex hyperbolic functions parameterize the unit hyperbola $x^2 - y^2 = 1$ in the complex plane, mapping strip domains into entire complex planes or Riemann surfaces due to their $2\pi i$ periodicity.

⚙️ Algebraic & Structural Meaning

They establish a rigorous analytical isomorphism between trigonometric functions of real/imaginary arguments and hyperbolic functions via Gudermannian and Euler connections, notably $\sin(iz) = i\sinh(z)$ and $\cos(iz) = \cosh(z)$.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Holomorphicity / Entirety	`$z \in \mathbb{C}$ (Entire complex plane)`	Strictly Mandatory / Necessary	Unlike real inverse hyperbolic functions which have restricted domains, $\sinh(z)$ and $\cosh(z)$ are entire functions with no singularities in the finite complex plane.
Periodicity in Imaginary Direction	`Period $T = 2\pi i$ for sine/cosine`	Strictly Mandatory	Ignoring periodicity leads to loss of infinite solutions when solving transcendental equations like $\sinh(z) = k$.
Branch Cut Restrictions for Inverses	`Domain boundaries for $\sinh^{-1}(z)$ and $\cosh^{-1}(z)$`	Essential for principal value determination	Failure to account for logarithmic branch cuts results in multivalued error selections in multiple-choice questions.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Separation into Real and Imaginary Parts

$$ \sinh(x + iy) = \sinh x \cos y + i \cosh x \sin y $$

Rule: Direct expansion rule using addition theorems for hyperbolic and circular functions. Memorize $\sinh(x+iy) = \sinh x \cos y + i \cosh x \sin y$ and $\cosh(x+iy) = \cosh x \cos y + i \sinh x \sin y$.

Application: undefined

Link Between Circular and Hyperbolic Functions

$$ \sin(iz) = i \sinh(z), \quad \cos(iz) = \cosh(z), \quad \tan(iz) = i \tanh(z) $$

Rule: Instantly convert complex circular function questions into hyperbolic functions or vice versa by substituting $z$ with $iz$.

Logarithmic Form of Inverse Hyperbolic Sine

$$ \sinh^{-1}(z) = \log\left(z + \sqrt{z^2 + 1}\right) $$

Rule: Convert inverse hyperbolic equations into standard logarithmic equations to solve for $z$.

Application: Standard direct formula question in UP TGT / PGT Mathematics.

Logarithmic Form of Inverse Hyperbolic Cosine

$$ \cosh^{-1}(z) = \log\left(z + \sqrt{z^2 - 1}\right) $$

Rule: Use the radical and logarithmic conversion directly when evaluating complex inverse values.

Application: Tested in DSSSB PGT Mathematics for determining real and imaginary parts of inverse values.

Modulus Square of Hyperbolic Sine

$$ |\sinh(x + iy)|^2 = \sinh^2 x + \sin^2 y $$

Rule: Sum of squared real hyperbolic and squared real trigonometric parts gives the magnitude squared instantly.

Application: Crucial for magnitude and inequality problems in LT Grade and UP PGT.

Rapid Elimination via Periodicity

$$ f(z + 2\pi i) = f(z) \text{ for } \sinh(z), \cosh(z) $$

Rule: Eliminate options that lack the correct periodic shift ($2\pi i$ for sine/cosine, $\pi i$ for tangent).

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Confusing Real Hyperbolic Boundedness with Complex Boundedness

Common Pitfall: Students assume $\sinh(z)$ is bounded because real $\sinh(x)$ grows monotonically. In $\mathbb{C}$, $\sinh(z)$ takes all complex values infinitely often.

Counterexample: \sinh(x + iy) = k \text{ has infinitely many solutions in } \mathbb{C}

Why it fails: Because of the periodic factor $\sin y$ and $\cos y$, complex hyperbolic functions are surjective onto $\mathbb{C}$ (except $\cosh z = \pm 1$ has specific root structures), unlike their real counterparts.

🛡️ Exam Defense: Remember that complex trigonometric and hyperbolic functions are unbounded in modulus over the complex plane.

⚠️ COMMON ERROR #02

Sign Errors in Real/Imaginary Expansion

Common Pitfall: Mixing up the plus and minus signs or placing $i$ incorrectly in $\cosh(x+iy) = \cosh x \cos y + i \sinh x \sin y$.

Counterexample: \cosh(x+iy) \neq \cosh x \cos y - i \sinh x \sin y

Why it fails: Expanding via $e^{x+iy}$ yields $\cosh x \cos y + i \sinh x \sin y$. A sign error ruins the entire derivation.

🛡️ Exam Defense: Verify using $z = ix$: $\cosh(ix) = \cos(x)$, which matches $\cosh(0)\cos(x) + i\sinh(0)\sin(x) = \cos(x)$. Always check with $x=0$ or $y=0$.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Complex Sine/Cosine	$z \in \mathbb{C}$	$\sin(z) = \sin x \cosh y + i \cos x \sinh y$	Maps Cartesian grids to elliptic coordinates	Keywords: 'Real part of $\sin(x+iy)$'
Complex Hyperbolic	$z \in \mathbb{C}$	$\sinh(x+iy) = \sinh x \cos y + i \cosh x \sin y$	Parameterizes hyperbolas in phase space	Keywords: 'Value of $\cosh(x+iy)$'
Inverse Hyperbolic Sine	$z \in \mathbb{C}$	$\sinh^{-1}(z) = \log(z + \sqrt{z^2 + 1})$	Logarithmic representation of inverse mapping	Keywords: 'Express $\sinh^{-1}(x+iy)$ in log form'
Periodicity Relation	$z \in \mathbb{C}$	Period is $2\pi i$ for $\sinh$, $\pi i$ for $\tanh$	Repetitive strip mapping across imaginary axis	Keywords: 'General solution of $\cosh z = 0$'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If $\sin(x + iy) = A + iB$, what is the value of $A^2 \sec^2 x - B^2 \csc^2 x$?

(A) $1$(B) $-1$(C) $0$(D) $2$

📚 Method A: Formal Textbook Derivation

Step 1: Expand $\sin(x+iy) = \sin x \cosh y + i \cos x \sinh y$. Step 2: Equate real and imaginary parts: $A = \sin x \cosh y$, $B = \cos x \sinh y$. Step 3: Express $\sec^2 x = \frac{1}{\cos^2 x}$ and $\csc^2 x = \frac{1}{\sin^2 x}$. Step 4: Compute $A^2 \sec^2 x - B^2 \csc^2 x = \frac{\sin^2 x \cosh^2 x}{\cos^2 x} - \frac{\cos^2 x \sinh^2 x}{\sin^2 x} = \tan^2 x \cosh^2 x - \cot^2 x \sinh^2 x$. Using $\cosh^2 y = 1 + \sinh^2 y$, simplify to $1$.

⚡ Method B: 25-Second Exam Shortcut

Put $y = 0$. Then $\sin x = A$, $B = 0$. Expression becomes $A^2 \cdot \frac{1}{\cos^2 x} = \sin^2 x \cdot \frac{1}{\cos^2 x} = \tan^2 x$. Wait, test with specific values: if $x = \pi/4, y = 0$, $A = 1/\sqrt{2}, B = 0$, yielding $1$.

💡 Tip: This is a classic UP PGT repeating identity. Memorize the result as identically $1$.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Separate into real and imaginary parts: $\tan(x + iy)$.

(A) $\frac{\sin 2x + i \sinh 2y}{\cos 2x + \cosh 2y}$(B) $\frac{\sin 2x - i \sinh 2y}{\cos 2x + \cosh 2y}$(C) $\frac{\sinh 2x + i \sin 2y}{\cosh 2x + \cos 2y}$(D) $\frac{\sin 2x + i \cosh 2y}{\cos 2x + \sinh 2y}$

📚 Method A: Formal Textbook Derivation

Step 1: Write $\tan(x+iy) = \frac{\sin(x+iy)}{\cos(x+iy)}$. Step 2: Multiply numerator and denominator by the complex conjugate of the denominator $\cos(x-iy)$. Step 3: Use product-to-sum trigonometric and hyperbolic identities to simplify numerator to $\sin 2x + i \sinh 2y$ and denominator to $\cos 2x + \cosh 2y$.

⚡ Method B: 25-Second Exam Shortcut

Check denominator: denominator for $\tan(x+iy)$ expansion always involves $\cos 2x + \cosh 2y$. Only option (A) has $\sin 2x + i \sinh 2y$ over $\cos 2x + \cosh 2y$.

💡 Tip: Denominator is invariably $\cos 2x + \cosh 2y$ for both tangent and cotangent complex expansions.

Practice questions

Choose an answer to check it and see the solution:

What is the value of $\cosh(i\pi)$?

A $1$

B $-1$

C $i$

D $-i$

Derivation & Analysis:

Using $\cosh(iz) = \cos(z)$, we have $\cosh(i\pi) = \cos(\pi) = -1$. Alternatively, $\frac{e^{i\pi} + e^{-i\pi}}{2} = \frac{-1 + (-1)}{2} = -1$.

The real part of $\sin(x + iy)$ is:

A $\sin x \cos y$

B $\sin x \cosh y$

C $\cos x \sinh y$

D $\sinh x \sin y$

Derivation & Analysis:

$\sin(x+iy) = \sin x \cos(iy) + \cos x \sin(iy) = \sin x \cosh y + i \cos x \sinh y$. Thus, the real part is $\sin x \cosh y$.

If $\cosh z = 0$, then $z$ equals:

A $(2n + 1)\pi i$

B $(2n + 1)\frac{\pi}{2} i$

C $n\pi i$

D $2n\pi i$

Derivation & Analysis:

$\cosh z = \cos(iz) = 0 \implies iz = (2n + 1)\frac{\pi}{2} \implies z = (2n + 1)\frac{\pi}{2}i$.

What is the value of $\sinh^{-1}(i)$?

A $\frac{\pi}{2}i$

B $i \log(1+\sqrt{2})$

C $\pm \frac{\pi}{2}i$

D $(\frac{\pi}{2} + 2n\pi)i$

Derivation & Analysis:

Let $z = \sinh^{-1}(i) \implies \sinh z = i \implies \frac{e^z - e^{-z}}{2} = i \implies e^z - 2ei - e^{-z} = 0$. Solving gives $z = \frac{\pi}{2}i$ (principal value).

The magnitude $|\cosh(x + iy)|^2$ is given by:

A $\sinh^2 x + \sin^2 y$

B $\cosh^2 x - \sin^2 y$

C $\cosh^2 x + \sinh^2 y$

D $\cos^2 x + \cosh^2 y$

Derivation & Analysis:

$|\cosh(x+iy)|^2 = \cosh^2 x \cos^2 y + \sinh^2 x \sin^2 y = \cosh^2 x (1 - \sin^2 y) + (\cosh^2 x - 1)\sin^2 y = \cosh^2 x - \sin^2 y$.

If $\tan(x + iy) = u + iv$, then $u$ is equal to:

A $\frac{\sin 2x}{\cos 2x + \cosh 2y}$

B $\frac{\sinh 2x}{\cosh 2x + \cos 2y}$

C $\frac{\sin 2x}{\cos 2x - \cosh 2y}$

D $\frac{\sinh 2x}{\cosh 2x - \cos 2y}$

Derivation & Analysis:

Rationalizing $\frac{\sin(x+iy)}{\cos(x+iy)}$ yields the real part $u = \frac{\sin 2x}{\cos 2x + \cosh 2y}$.

Which of the following is identically true for all complex $z$?

A $\cosh^2 z - \sinh^2 z = 1$

B $\cosh^2 z + \sinh^2 z = 1$

C $\tanh^2 z + \sech^2 z = 1$

D Both (A) and (C)

Derivation & Analysis:

Both hyperbolic identity variations $\cosh^2 z - \sinh^2 z = 1$ and $\tanh^2 z + \sech^2 z = 1$ hold analytically for all complex $z$ where defined.

The value of $\tan(i \log(\alpha))$ where $\alpha > 0$ is:

A $i \frac{\alpha^2 - 1}{\alpha^2 + 1}$

B $\frac{\alpha^2 - 1}{\alpha^2 + 1}$

C $i \frac{\alpha - 1}{\alpha + 1}$

D $\frac{\alpha - 1}{\alpha + 1}$

Derivation & Analysis:

$\tan(iz) = i \tanh(z)$. Here $z = \log \alpha$. Thus $i \tanh(\log \alpha) = i \frac{e^{\log \alpha} - e^{-\log \alpha}}{e^{\log \alpha} + e^{-\log \alpha}} = i \frac{\alpha - \alpha^{-1}}{\alpha + \alpha^{-1}} = i \frac{\alpha^2 - 1}{\alpha^2 + 1}$.

For what values of $z$ does $\sinh z = 0$?

A $z = n\pi$

B $z = n\pi i$

C $z = (2n+1)\frac{\pi}{2}$

D $z = 0$

Derivation & Analysis:

$\sinh z = \frac{e^z - e^{-z}}{2} = 0 \implies e^{2z} = 1 = e^{2n\pi i} \implies 2z = 2n\pi i \implies z = n\pi i$.

Q10

If $\log(\sin(x + iy)) = a + ib$, then $e^{2a}$ equals:

A $\sinh^2 x + \cos^2 y$

B $\cosh^2 x - \sin^2 y$

C $\sin^2 x + \sinh^2 y$

D $\cos^2 x + \sinh^2 y$

Derivation & Analysis:

Taking exponential of both sides, $e^{a+ib} = \sin(x+iy) \implies e^a = |\sin(x+iy)| = \sqrt{\sin^2 x + \sinh^2 y}$. Squaring gives $e^{2a} = \sin^2 x + \sinh^2 y$.

Common questions

Yes. $\sinh(z)$ and $\cosh(z)$ are periodic with purely imaginary period $2\pi i$. $\tanh(z)$ has period $\pi i$. This is a direct consequence of their exponential definitions combined with the $2\pi$-periodicity of the exponential function.

Students mistakenly apply circular trigonometric expansion rules directly without accounting for $\cos(iy) = \cosh(y)$ and $\sin(iy) = i\sinh(y)$. Always substitute $z = x + iy$ into $\frac{e^z + e^{-z}}{2}$ to derive expansions from first principles during exams.

Yes! Because the imaginary argument introduces $\cosh y$ and $\sinh y$ factors which grow exponentially, complex sine and cosine can take any complex value whatsoever, destroying the real $[-1, 1]$ bounds.

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