Problem 59 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\sin ^{-1}\left[\frac{1}{i}(z-1)\right]\), where \(z\) is non-real, can be the angle of a triangle if (A) \(\operatorname{Re}(z)=1, \operatorname{Im}(z)=2\) (B) \(\operatorname{Re}(z)=1,-1 \leq \operatorname{Im}(z) \leq 1\) (C) \(\operatorname{Re}(z)+\operatorname{Im}(z)=0\) (D) \(\operatorname{Re}(z)=\operatorname{Im}(z)\)

Step-by-Step Solution

Answer

Option C is correct: \(\operatorname{Re}(z) + \operatorname{Im}(z) = 0\).

Step 1: Substitute the given condition into the expression

We need to find if \( \sin^{-1}\left[\frac{1}{i}(z-1)\right] \) can be an angle of a triangle. This means the inverse sine function should have a real value in \([0, \pi]\). Substitute \(z = x + yi\) into \(\frac{1}{i}(z-1)\): \[ \frac{1}{i}(x + yi - 1) = -i(x + yi - 1) = y + (1-x)i \].

Step 2: Analyze when the complex inverse sine is an angle of a triangle

For \( \sin^{-1}(y + (1-x)i) \) to be an angle in a triangle, \(y + (1-x)i\) must be a purely imaginary number such that its modulus is \(\leq 1\). Thus, \(|y| \leq 1-x\).

Step 3: Evaluate options based on conditions

Check each condition:\- **Option (A):** \(\operatorname{Re}(z)=1, \operatorname{Im}(z)=2\) \( z - 1 = yi = 2i \), which doesn't satisfy \(|2| \leq 1\).- **Option (B):** \(\operatorname{Re}(z)=1,-1 \leq \operatorname{Im}(z) \leq 1\) \( z - 1 = yi, \) but since \(x = 1\), 0 must be \(>\) modulus condition. Doesn't satisfy.- **Option (C):** Check if \(\operatorname{Re}(z) + \operatorname{Im}(z) = 0\), (imaginary valid constraint) \(\Rightarrow x = y\) makes real part zero, confirming imaginary.\- **Option (D):** \(\operatorname{Re}(z) = \operatorname{Im}(z)\), \(z - 1 = i(y-y)\), results real values possibly forming angles, invalid here.

Step 4: Choose the correct option

The conditions must ensure \( (1-x) \) is equal to zero or satisfied. Only option C aligns with forming a pure condition ideas, correctly \(|y|\leq0\). Thus, \(C\) satisfies.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverses of the trigonometric functions. They are used to determine angles when you know the trigonometric ratio. For example, if you know the sine of an angle, the inverse sine function, denoted \( \sin^{-1} \), will give you the angle. In this exercise, we're dealing with the inverse sine of a complex number, \( \sin^{-1}[y + (1-x)i] \). This function will yield a real angle if the input is a purely imaginary number with a modulus less than or equal to one. This is key in ensuring the result is a valid angle in a triangle, which can only be between \([0, \pi]\). To find such conditions, we analyze the modulus which must satisfy \(|y| \leq 1-x\).

Inverse trigonometric functions can seem complex when extended to complex numbers, but they are immensely useful because they make it possible to work with non-real numbers in solving geometric problems. The key takeaway is that understanding if a complex number can represent a valid angle involves ensuring it fits within the expected real trigonometric value range.

Trigonometry in Triangles

Trigonometry in triangles involves the relationship between the angles and sides of triangles. It is crucial for understanding and solving many problems in math and engineering. When analyzing if a complex number can represent an angle of a triangle, we apply trigonometric concepts. A primary condition for this is that measured angles need to be real and between \(0\) and \(\pi\) radians.

In our exercise, we investigated whether \( \sin^{-1}(y+(1-x)i) \) could be one of these angles. By expressing the complex number in terms of its real and imaginary parts, we found conditions under which the complex number could serve as a valid angle. Specifically, we explored when this expression becomes purely imaginary and satisfies the requirement \(|y| \leq 1-x\).

This scenario demonstrates the fusion of algebraic manipulation with trigonometric principles in complex numbers. Understanding such exercises helps solidify the foundational concept that angles in a triangle must remain within feasible bounds, a fundamental necessity in trigonometry.

Complex Modulus

The complex modulus is a vital concept when dealing with complex numbers. For a complex number \( z = x + yi \), its modulus is given by \( |z| = \sqrt{x^2 + y^2} \). It measures the "length" of the vector from the origin to the point \( (x, y) \) in the complex plane. Understanding modulus is essential when determining if an expression involving complex numbers can yield specific trigonometric values.

In the context of our exercise, we find that the modulus condition \(|y| \leq 1-x\) is derived by ensuring that the expression inside the inverse trigonometric function is of a specific form: purely imaginary and with a modulus less than or equal to one. This is because, in trigonometric terms, it translates to an angle's sine being a valid real number between -1 and 1.

The modulus helps not only in calculating lengths and distances in the complex plane but also provides a way to verify conditions when determining if a complex number meets the criteria for being a possible angle of a triangle. It is a bridge that connects algebraic expressions to geometric and trigonometric interpretations.