Complex Numbers

Circular functions of a complex variable

Circular functions of a complex variable extend standard real trigonometric functions via Euler's formula, establishing deep connections with hyperbolic functions.

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

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Definition and statement

Euler's Definitions & Addition Theorems for Complex Circular Functions

For any complex number $z = x + iy$, where $x, y \in \mathbb{R}$, the complex sine, cosine, and tangent are defined as $^\dagger \sin z = \frac{e^{iz} - e^{-iz}}{2i}$, $\cos z = \frac{e^{iz} + e^{-iz}}{2}$, and $\tan z = \frac{\sin z}{\cos z}$, satisfying fundamental periodicity and hyper-trigonometric transformations.

📐 Geometric Interpretation

While real circular functions map real intervals to bounded ranges $[-1, 1]$, complex circular functions map the complex plane (Cartesian strip domains) to unbounded regions in $\mathbb{C}$, bridging Euclidean rotation with hyperbolic scaling.

⚙️ Algebraic & Structural Meaning

They obey analogous identities to real trigonometry (e.g., $\sin^2 z + \cos^2 z = 1$), but exhibit unbounded moduli for purely imaginary arguments due to the underlying exponential growth of $\sinh y$ and $\cosh y$.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Analyticity on $\mathbb{C}$	`$z \in \mathbb{C}$ (Entire domain)`	Strictly Mandatory / Necessary	For functions like $\sec z$ or $\csc z$, singularities occur where $\cos z = 0$ or $\sin z = 0$, leading to poles and non-analyticity.
Periodicity domain	$z \in \mathbb{C}$	Mandatory for periodic property mapping	Failure to account for $2\pi$ periodicity along the real axis leads to loss of infinite solution branches.
Hyperbolic argument separation	`Cartesian splitting $z = x + iy$`	Essential for real and imaginary part extraction	Directly treating complex variables as real variables causes severe sign errors, especially with $i^2 = -1$ in hyperbolic conversions.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Real and Imaginary Parts of Complex Sine

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

Rule: Memorize the sign and hyperbolic alternation: Sine product uses $\sin x \cosh y + i \cos x \sinh y$.

Application: Frequently asked in UP PGT to find the real or imaginary part of $\sin(x+iy)$ or $\cos(x+iy)$.

Real and Imaginary Parts of Complex Cosine

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

Rule: Note the minus sign for cosine: $\cos x \cosh y - i \sin x \sinh y$.

Application: Direct matching options in TGT/PGT screening tests.

Modulus of Complex Sine and Cosine

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

Rule: Use $\cosh^2 y - \sinh^2 y = 1$ to quickly evaluate squared moduli without expanding complex exponentials.

Application: Standard numerical evaluation questions in KVS/DSSSB PGT.

Purely Imaginary Argument Reduction

$$\sin(iy) = i \sinh y, \quad \cos(iy) = \cosh y, \quad \tan(iy) = i \tanh y$$

Rule: Circular function of pure imaginary number converts directly into hyperbolic function with an $i$ factor for sine/tangent.

Application: Rapid elimination trick for multi-statement assertion questions.

Magnitude Bounds for Sine/Cosine

$$|\sin z| \ge |\sin x|, \quad |\cos z| \ge |\cos x| \text{ when } y \neq 0$$

Rule: Since $\cosh y \ge 1$ for all real $y$, the modulus can exceed 1 unlike real trigonometry.

Application: Tested in conceptual statement-based questions.

Periodicity of Complex Tangent

$$\tan(z + \pi) = \tan z, \quad \text{Fundamental period is } \pi$$

Rule: Tangent and cotangent maintain period $\pi$, while sine and cosine maintain $2\pi$ in the complex plane.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Boundedness Fallacy

Common Pitfall: Assuming $|\sin z| \le 1$ for all complex numbers $z$, carrying over real trigonometric intuition.

Counterexample: $z = i y$, where $\sin(iy) = i \sinh y \to \infty$ as $y \to \infty$

Why it fails: Hyperbolic sine grows exponentially; hence complex sine and cosine are unbounded in the complex plane.

🛡️ Exam Defense: Never select options stating that complex circular functions are bounded on $\mathbb{C}$.

⚠️ COMMON ERROR #02

Hyperbolic Sign Confusion

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

Counterexample: Using $+i$ instead of $-i$ in the expansion of $\cos(x+iy)$

Why it fails: Leads to incorrect conjugate matching and wrong sign in options (A) vs (B).

🛡️ Exam Defense: Remember: Sine preserves the inner sign ($+$), while Cosine reverses it ($-$).

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Real Sine function	$x \in \mathbb{R}$	$\sin x \in [-1, 1]$	Projection on vertical unit circle axis	Bounded range problems
Complex Sine function	$z = x+iy \in \mathbb{C}$	$\sin x \cosh y + i \cos x \sinh y$	Maps vertical strips to elliptic regions / unbounded	Involves $\sinh$ and $\cosh$ terms
Pure Imaginary Sine	$z = iy$	$\sin(iy) = i \sinh y$	Purely imaginary vertical growth	Argument contains only $i$
Modulus Squared Identity	$z = x+iy$	$\|\sin z\|^2 = \sin^2 x + \sinh^2 y$	Square distance from origin in range space	Asks for magnitude or real value evaluation

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If $\sin(x + iy) = \u + iv$, find the value of $u$ and $v$ in terms of $x$ and $y$.

(A) $u = \sin x \cosh y, v = \cos x \sinh y$(B) $u = \sin x \sinh y, v = \cos x \cosh y$(C) $u = \cos x \cosh y, v = \sin x \sinh y$(D) $u = \cos x \sinh y, v = \sin x \cosh y$

📚 Method A: Formal Textbook Derivation

Step 1: Write $\sin(x+iy) = \sin x \cos(iy) + \cos x \sin(iy)$. Step 2: Use $\cos(iy) = \cosh y$ and $\sin(iy) = i \sinh y$. Step 3: Equate real ($u$) and imaginary ($v$) parts to get $u = \sin x \cosh y$ and $v = \cos x \sinh y$.

⚡ Method B: 25-Second Exam Shortcut

Recall standard expansion formula: Sine pairs $\sin x$ with $\cosh y$ for real part and $\cos x$ with $\sinh y$ for imaginary part.

💡 Tip: Direct formula recall item. Do not waste time deriving during exam.

CANONICAL PROBLEM 02 Dual-Method Breakdown

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

(A) $0$(B) $1$(C) $-1$(D) $\cosh^2 y$

📚 Method A: Formal Textbook Derivation

Step 1: Here $A = \cos x \cosh y$ and $B = -\sin x \sinh y$. Step 2: Compute $A^2 = \cos^2 x \cosh^2 y$ and $B^2 = \sin^2 x \sinh^2 y$. Step 3: Substitute into expression using $\cosh^2 y - \sinh^2 y = 1$.

⚡ Method B: 25-Second Exam Shortcut

Recognize standard identity transformation yielding unity.

💡 Tip: Look for $\cosh^2 y - \sinh^2 y = 1$ hidden inside trigonometric identities.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: The value of $\sin\left(\frac{\pi}{2} + i\log 2\right)$ is equal to:

A $\frac{5}{4}$

B $\frac{5}{8}$

C $\frac{3}{4}$

D $2$

Derivation & Analysis:

Using $\sin(\frac{\pi}{2} + iy) = \sin(\frac{\pi}{2}) \cosh y = \cosh y$. Here $y = \log 2$, so $\cosh(\log 2) = \frac{e^{\log 2} + e^{-\log 2}}{2} = \frac{2 + 1/2}{2} = \frac{5}{4}.

MCQ 2: If $\tan(x + iy) = u + iv$, then $u$ is given by:

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

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

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

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

Derivation & Analysis:

Expanding $\tan(x+iy) = \frac{\sin(x+iy)}{\cos(x+iy)}$ and multiplying numerator and denominator by $\cos(x-iy)$ yields $u = \frac{\sin 2x}{\cosh 2y + \cos 2x}$.

MCQ 3: The real part of $\cos(i)$ is:

A $\cos 1$

B $\cosh 1$

C $-\cosh 1$

D $i \sinh 1$

Derivation & Analysis:

Since $\cos(iy) = \cosh y$, for $y = 1$, $\cos(i) = \cosh 1$, which is real.

MCQ 4: Which of the following relations is correct for complex variables?

A $\sin^2 z + \cos^2 z = 1$

B $\sin^2 z - \cos^2 z = 1$

C $1 + \tan^2 z = \csc^2 z$

D $1 - \tan^2 z = \sec^2 z$

Derivation & Analysis:

Standard trigonometric identity $\sin^2 z + \cos^2 z = 1$ holds identically for all complex $z$ due to analytic continuation.

MCQ 5: If $\sec(x + iy) = u + iv$, then $v$ is equal to:

A $\sec x \sech y$

B $\sin x \tanh y$

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

D $\frac{2 \cos x \cosh y}{\cos 2x + \cosh 2y}$

Derivation & Analysis:

Since $\sec(x+iy) = \frac{1}{\cos x \cosh y - i \sin x \sinh y}$, rationalizing gives the imaginary part as $\frac{2 \sin x \sinh y}{\cos 2x + \cosh 2y}$.

MCQ 6: The value of $|\sin(x+iy)|^2$ is:

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

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

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

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

Derivation & Analysis:

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

MCQ 7: If $\alpha + i\beta = \log(\sin(x + iy))$, then $\beta$ equals:

A $\tan^{-1}\left(\frac{\cot x}{\tanh y}\right)$

B $\tan^{-1}\left(\frac{\tan x}{\tanh y}\right)$

C $\tan^{-1}\left(\frac{\tanh y}{\tan x}\right)$

D $\tan^{-1}(\tan x \tanh y)$

Derivation & Analysis:

Taking imaginary parts of $\log(\sin(x+iy))$ gives $\beta = \arg(\sin(x+iy)) = \tan^{-1}\left(\frac{\cos x \sinh y}{\sin x \cosh y}\right) = \tan^{-1}\left(\frac{\cot x}{\tanh y}\right)$, wait let's check: $\frac{v}{u} = \frac{\cos x \sinh y}{\sin x \cosh y} = \cot x \tanh y = \frac{\tanh y}{\tan x}$. Therefore $\beta = \tan^{-1}\left(\frac{\tanh y}{\tan x}\right)$.

MCQ 8: For a complex number $z$, $\sin z = 0$ implies:

A $z = n\pi, n \in \mathbb{Z}$

B $z = i n\pi, n \in \mathbb{Z}$

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

D $z = 0$

Derivation & Analysis:

$\sin z = \frac{e^{iz} - e^{-iz}}{2i} = 0 \implies e^{2iz} = 1 \implies 2iz = 2n\pi i \implies z = n\pi$ for $n \in \mathbb{Z}$. Roots remain purely real.

MCQ 9: The value of $\cosh(i\pi)$ is:

A $1$

B $-1$

C $0$

D $i$

Derivation & Analysis:

$\cosh(i\pi) = \cos(\pi) = -1$ using the relation $\cosh(iy) = \cos y$.

Q10

MCQ 10: If $\sin z = x$, and $x > 1$ is real, then $z$ is:

A Purely real

B Purely imaginary

C Complex with real part $\frac{\pi}{2}$

D Complex with imaginary part $0$

Derivation & Analysis:

Solving $\sin z = x > 1$ yields $z = \frac{\pi}{2} + i y$ or similar branches where the real part is an odd multiple of $\frac{\pi}{2}$.

Common questions

Yes! Unlike real trigonometry where $-1 \le \sin x \le 1$, the complex sine function contains hyperbolic terms like $\sinh^2 y$, which grow without bound as $y \to \infty$. Thus, $|\sin z|$ can take any real value greater than or equal to $|\sin x|$. voulu.

Confusing signs in product expansions (such as $\cos(x+iy) = \cos x \cosh y - i \sin x \sinh y$) and forgetting that $\sin(iy) = i \sinh y$ contains an extra factor of $i$.

Memorize standard expansion formulas and reduction rules for $iy$ arguments to bypass full algebraic derivation.

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