Q.The principal value of $\tan^{-1}\sqrt3$ is __________.

Showing the 12 most recent of 20 on this concept.

1. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.The amplitude of the complex number $\dfrac{(\sqrt{3}+i)(1-\sqrt{3}i)}{(-1+i)(-1-i)}$ is (A) $\dfrac{\pi}{2}$ (B) $\dfrac{\pi}{3}$ (C) $-\dfrac{5\pi}{12}$ (D) $-\dfrac{\pi}{6}$

   ›Reveal solutionSolution

   The amplitude (argument) is $-\dfrac{\pi}{6}$, so the answer is (D).

   Write each factor in polar form and combine arguments (add for products, subtract for the quotient).

   Numerator

   • $\sqrt{3}+i$: modulus $2$, argument $\dfrac{\pi}{6}$.
   • $1-\sqrt{3}\,i$: modulus $2$, argument $-\dfrac{\pi}{3}$.

   Numerator argument $=\dfrac{\pi}{6}-\dfrac{\pi}{3}=-\dfrac{\pi}{6}$.

   Denominator …

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2. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.The amplitude of the complex number $\dfrac{(\sqrt{3}+i)(1-\sqrt{3}i)}{(-1+i)(-1-i)}$ is (A) $\dfrac{\pi}{2}$ (B) $\dfrac{\pi}{3}$ (C) $-\dfrac{5\pi}{12}$ (D) $-\dfrac{\pi}{6}$

   ›Reveal solutionSolution

   The expression reduces to $\sqrt{3}-i$, whose amplitude is $-\dfrac{\pi}{6}$. Option (D).

   Solution

   Simplify numerator and denominator directly.

   Numerator:

   $(\sqrt{3}+i)(1-\sqrt{3}\,i)=\sqrt{3}-3i+i-\sqrt{3}\,i^{2}=\sqrt{3}-2i+\sqrt{3}=2\sqrt{3}-2i.$

   Denominator: the factors are conjugates, so

   $(-1+i)(-1-i)=(-1)^{2}-(i)^{2}=1+1=2.$

   Therefore

   $z=\frac{2\sqrt{3}-2i}{2}=\sqrt{3}-i.$ …

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3. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.If $z = \dfrac{(2-i)(1+i)^3}{(1-i)^2}$, then Arg$(z)=$ (A) $\tan^{-1}\left(\dfrac{1}{3}\right) - \pi$ (B) $\tan^{-1}\left(\dfrac{3}{4}\right) - \pi$ (C) $\pi - \tan^{-1}\left(\dfrac{3}{4}\right)$ (D) $\tan^{-1}\left(\dfrac{1}{3}\right)$

   ›Reveal solutionSolution

   The key idea is to simplify the complex number $z$ into standard form $a+bi$ by expanding powers and rationalising, then compute its argument using $\tan^{-1}(b/a)$ and adjust for the correct quadrant. The final argument is $\tan^{-1}(1/3) - \pi$, so the correct option is (A).

   We are given

   $z = \frac{(2-i)(1+i)^3}{(1-i)^2}.$

   We need $\operatorname{Arg}(z)$, the principal argument (usually in $(-\pi, \pi]$). The trick is to simplify $z$ to a form $a+bi$ and then find the angle.

   ---

   Concept & Intuition

   The argument of a complex number is the angle it makes with the positive real axis. When a complex number is expressed as a fraction of products, we can either simplify algebraically or use properties of arguments (sum/difference of angles). Here, direct algebraic simplification is straightforward because the powers are small. The pitfall: forgetting to check which quadrant the simplified number lies in — the arctan formula alone gives only the reference angle.

   ---

   Step-by-step solution

   1. Simplify $(1+i)^3$

      First, $(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$.

      Then $(1+i)^3 = (1+i)(2i) = 2i + 2i^2 = 2i - 2 = -2 + 2i$.

   2. Simplify $(1-i)^2$

      $(1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i$.

   3. Write $z$ with these simplifications

   $z = \frac{(2-i)(-2 + 2i)}{-2i}.$

   4. Multiply numerator

      $(2-i)(-2+2i) = 2(-2+2i) - i(-2+2i) = -4 + 4i + 2i - 2i^2$.

      Since $i^2 = -1$, $-2i^2 = -2(-1) = 2$.

      So numerator = $-4 + 4i + 2i + 2 = -2 + 6i$.

   5. Now $z = \dfrac{-2 + 6i}{-2i}$

      Multiply numerator and denominator by $i$ to rationalise:

   $z = \frac{(-2+6i) \cdot i}{-2i \cdot i} = \frac{-2i + 6i^2}{-2i^2}.$

   Since $i^2 = -1$, we get:

   $z = \frac{-2i + 6(-1)}{-2(-1)} = \frac{-2i - 6}{2} = \frac{-6 - 2i}{2} = -3 - i.$

   6. Identify real and imaginary parts

      So $z = -3 - i$, i.e. $a = -3$, $b = -1$. Both are negative, so $z$ lies in the third quadrant.

   7. Find the argument

      The reference angle $\theta_0 = \tan^{-1}\left|\frac{b}{a}\right| = \tan^{-1}\left(\frac{1}{3}\right)$. …

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4. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If $f(x) = \tan\left(\dfrac{\pi}{\sqrt{x+1}+4}\right)$ is a real valued function then the range of $f$ is (A) $[-1,1]$ (B) $(0,1]$ (C) $[-1,\infty)$ (D) $\mathbb{R}$

   ›Reveal solutionSolution

   The function is a composition of a decreasing inner square-root expression and the tangent function over a restricted domain; the range is $(0,1]$, so option (B) is correct.

   We need the range of $f(x) = \tan\left(\dfrac{\pi}{\sqrt{x+1}+4}\right)$ for real $x$ where the function is defined.

   The key is to find what values the argument of $\tan$ can take, then see what outputs $\tan$ gives on that interval.

   ---

   Concept & Intuition

   The tangent function is periodic and unbounded, but here its argument is squeezed into a small interval because the denominator $\sqrt{x+1}+4$ is always at least $4$. That means the fraction $\frac{\pi}{\text{big number}}$ stays small — specifically between $0$ and $\frac{\pi}{4}$. On $(0, \frac{\pi}{4})$, $\tan$ is positive, increasing, and bounded between $0$ and $1$. So the range is $(0,1]$.

   ---

   Step-by-step reasoning

   1. Domain of $f$

      We need $\sqrt{x+1}$ defined, so $x+1 \ge 0 \Rightarrow x \ge -1$.

      Also, the denominator $\sqrt{x+1}+4$ is never zero (it’s always $\ge 4$), so no further restriction.

      Domain: $[-1, \infty)$.

   2. Behavior of the inner expression

      Let $t = \sqrt{x+1}$. For $x \ge -1$, $t \ge 0$.

      Then the argument of $\tan$ is

   $\theta = \frac{\pi}{t+4}.$

   As $x$ increases from $-1$ to $\infty$, $t$ increases from $0$ to $\infty$.

   3. Range of $\theta$

      • When $x = -1$: $t = 0$, so $\theta = \frac{\pi}{4}$.
      • As $x \to \infty$: $t \to \infty$, so $\theta \to 0^+$. Since $t+4$ increases, $\theta$ decreases continuously. Hence $\theta \in (0, \frac{\pi}{4}]$.

   4. Apply $\tan$

      On the interval $(0, \frac{\pi}{4}]$, $\tan$ is:

      • Continuous and strictly increasing.
      • $\tan(0^+) = 0^+$ (approaches 0 from above). …

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5. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.Let $k > 0$ and $t = \text{Sech}^{-1}\left(\frac{1}{2}\right) - \text{Cosech}^{-1}\left(\frac{3}{k}\right)$. If $3e^t = 2 + \sqrt{3}$, then $k =$ (A) $2$ (B) $4$ (C) $3\sqrt{3}$ (D) $3\sqrt{2}$

   ›Reveal solutionSolution

   This problem requires converting inverse hyperbolic functions into their logarithmic forms, then simplifying the expression for $t$ and solving the given exponential equation. We find that $k=4$.

   The core idea here is to express the inverse hyperbolic functions in their equivalent logarithmic forms. This transforms the problem from one involving less familiar functions into a standard algebraic manipulation of logarithms and exponentials.

   Here are the relevant logarithmic identities for the inverse hyperbolic functions involved:

   For $0 < x \le 1$, $\text{Sech}^{-1}(x) = \log\left(\frac{1 + \sqrt{1 - x^2}}{x}\right)$.

   For $x > 0$, $\text{Cosech}^{-1}(x) = \log\left(\frac{1 + \sqrt{1 + x^2}}{x}\right)$.

   Let's apply these formulas step-by-step.

   1. Evaluate $\text{Sech}^{-1}\left(\frac{1}{2}\right)$: We use the formula for $\text{Sech}^{-1}(x)$ with $x = \frac{1}{2}$.

   $\text{Sech}^{-1}\left(\frac{1}{2}\right) = \log\left(\frac{1 + \sqrt{1 - \left(\frac{1}{2}\right)^2}}{\frac{1}{2}}\right)$

   First, calculate the term under the square root:
   

   $1 - \left(\frac{1}{2}\right)^2 = 1 - \frac{1}{4} = \frac{3}{4}$

   So, $\sqrt{1 - \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
   Substitute this back into the logarithm:
   

   $\text{Sech}^{-1}\left(\frac{1}{2}\right) = \log\left(\frac{1 + \frac{\sqrt{3}}{2}}{\frac{1}{2}}\right) = \log\left(\frac{\frac{2 + \sqrt{3}}{2}}{\frac{1}{2}}\right) = \log(2 + \sqrt{3})$

   2. Evaluate $\text{Cosech}^{-1}\left(\frac{3}{k}\right)$: We use the formula for $\text{Cosech}^{-1}(x)$ with $x = \frac{3}{k}$. Since $k > 0$, we know $\frac{3}{k} > 0$.

   $\text{Cosech}^{-1}\left(\frac{3}{k}\right) = \log\left(\frac{1 + \sqrt{1 + \left(\frac{3}{k}\right)^2}}{\frac{3}{k}}\right)$

   Simplify the term under the square root:
   

   $1 + \left(\frac{3}{k}\right)^2 = 1 + \frac{9}{k^2} = \frac{k^2 + 9}{k^2}$

   So, $\sqrt{1 + \left(\frac{3}{k}\right)^2} = \sqrt{\frac{k^2 + 9}{k^2}} = \frac{\sqrt{k^2 + 9}}{k}$ (since $k > 0$).
   Substitute this back into the logarithm:
   

   $\text{Cosech}^{-1}\left(\frac{3}{k}\right) = \log\left(\frac{1 + \frac{\sqrt{k^2 + 9}}{k}}{\frac{3}{k}}\right) = \log\left(\frac{\frac{k + \sqrt{k^2 + 9}}{k}}{\frac{3}{k}}\right) = \log\left(\frac{k + \sqrt{k^2 + 9}}{3}\right)$

   3. Substitute these into the expression for $t$: We are given $t = \text{Sech}^{-1}\left(\frac{1}{2}\right) - \text{Cosech}^{-1}\left(\frac{3}{k}\right)$.

   $t = \log(2 + \sqrt{3}) - \log\left(\frac{k + \sqrt{k^2 + 9}}{3}\right)$

   Using the logarithm property $\log a - \log b = \log\left(\frac{a}{b}\right)$:
   

   $t = \log\left(\frac{2 + \sqrt{3}}{\frac{k + \sqrt{k^2 + 9}}{3}}\right) = \log\left(\frac{3(2 + \sqrt{3})}{k + \sqrt{k^2 + 9}}\right)$

   4. Use the given equation $3e^t = 2 + \sqrt{3}$ to solve for $k$: First, isolate $e^t$:

   $e^t = \frac{2 + \sqrt{3}}{3}$

   Now, substitute the expression for $t$ we found:
   $$ e^{\log\left(\frac{3(2 + \sqrt{3})}{k + \sqrt{k^2 + 9}}\right)} = \frac{2 + \sqrt{3}}{3} $$ …
   

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6. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If one of the values of $\sqrt{-1-\sqrt{3}i}$ is $a+i\beta$, $\alpha < 0$ and $\beta > 0$, then $\alpha =$ (A) $-\dfrac{1}{\sqrt{2}}$ (B) $-\dfrac{\sqrt{3}}{\sqrt{2}}$ (C) $-\sqrt{3}$ (D) $-\sqrt{2}$

   ›Reveal solutionSolution

   $-1-\sqrt3\,i = 2\,\text{cis}\tfrac{4\pi}{3}$; its square roots are $\sqrt2\,\text{cis}\tfrac{2\pi}{3}$ and $\sqrt2\,\text{cis}\tfrac{5\pi}{3}$. The root with $\alpha<0,\beta>0$ gives $\alpha = -\tfrac{1}{\sqrt2}$.

   Write the radicand in polar form. For $-1 - \sqrt3\,i$: modulus $= \sqrt{1+3} = 2$, and the point $(-1,-\sqrt3)$ is in the third quadrant, so its argument is $\tfrac{4\pi}{3}$ (equivalently $-\tfrac{2\pi}{3}$):

   $-1 - \sqrt3\,i = 2\left(\cos\tfrac{4\pi}{3} + i\sin\tfrac{4\pi}{3}\right).$

   The square roots have modulus $\sqrt2$ and arguments $\tfrac{1}{2}\cdot\tfrac{4\pi}{3} = \tfrac{2\pi}{3}$ and $\tfrac{2\pi}{3} + \pi = \tfrac{5\pi}{3}$: …

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7. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If $\sqrt[3]{i} = \text{cis} \alpha$, $\alpha$ belongs to second quadrant and $\sqrt[3]{-i} = \text{cis} \beta$, $\beta$ belongs to third quadrant then $\text{cis} \alpha + \text{cis} \beta =$ (A) $\sqrt{3}$ (B) $i$ (C) $-i$ (D) $-\sqrt{3}$

   ›Reveal solutionSolution

   The cube roots of $i$ and $-i$ are found by De Moivre’s theorem; using the quadrant conditions picks the correct angles, and their sum simplifies to $-\sqrt{3}$.

   The problem asks for the sum of two complex numbers, each given in cis form (cis $\theta = \cos\theta + i\sin\theta$). The cube root of a complex number is not unique — there are three distinct cube roots. The quadrant condition for $\alpha$ and $\beta$ picks exactly one of those three for each. Once we have the correct angles, adding the two cis values is straightforward trigonometry.

   ---

   1. Find the cube roots of $i$. Write $i$ in polar form: $i = \text{cis}\, \frac{\pi}{2}$. By De Moivre’s theorem, the cube roots are

   $\text{cis}\left(\frac{\pi/2 + 2k\pi}{3}\right),\quad k = 0,1,2.$

   That gives the three angles:

   $\frac{\pi}{6},\quad \frac{\pi}{6} + \frac{2\pi}{3} = \frac{5\pi}{6},\quad \frac{\pi}{6} + \frac{4\pi}{3} = \frac{3\pi}{2}.$

   The second quadrant contains angles between $\frac{\pi}{2}$ and $\pi$. Among these, only $\frac{5\pi}{6}$ lies there.

   Hence $\alpha = \frac{5\pi}{6}$.

   2. Find the cube roots of $-i$. Write $-i = \text{cis}\left(-\frac{\pi}{2}\right)$ or equivalently $\text{cis}\,\frac{3\pi}{2}$. The cube roots are

   $\text{cis}\left(\frac{-\pi/2 + 2k\pi}{3}\right),\quad k = 0,1,2.$

   The three angles are:

   $-\frac{\pi}{6},\quad -\frac{\pi}{6} + \frac{2\pi}{3} = \frac{\pi}{2},\quad -\frac{\pi}{6} + \frac{4\pi}{3} = \frac{7\pi}{6}.$

   The third quadrant covers angles from $\pi$ to $\frac{3\pi}{2}$. Only $\frac{7\pi}{6}$ falls there.

   Hence $\beta = \frac{7\pi}{6}$. …

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8. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.If $Z_1 = \sqrt{3} + i\sqrt{3}$ and $Z_2 = \sqrt{3} + i$, and $\left(\dfrac{Z_1}{Z_2}\right)^{50} = x + iy$, then the point $(x, y)$ lies in (A) first quadrant (B) second quadrant (C) third quadrant (D) fourth quadrant

   ›Reveal solutionSolution

   Work in polar form: $\arg(Z_1/Z_2)=\tfrac{\pi}{4}-\tfrac{\pi}{6}=\tfrac{\pi}{12}$, so the 50th power has argument $\tfrac{50\pi}{12}=\tfrac{25\pi}{6}\equiv\tfrac{\pi}{6}$. Both $x$ and $y$ are positive, so $(x,y)$ is in the first quadrant — option (A).

   Step 1 — argument of $Z_1=\sqrt3+i\sqrt3$.

   $\theta_1=\arctan\!\frac{\sqrt3}{\sqrt3}=\arctan 1=\frac{\pi}{4}.$

   Step 2 — argument of $Z_2=\sqrt3+i$.

   $\theta_2=\arctan\!\frac{1}{\sqrt3}=\frac{\pi}{6}.$

   Step 3 — argument of the quotient.

   $\arg\!\left(\frac{Z_1}{Z_2}\right)=\theta_1-\theta_2=\frac{\pi}{4}-\frac{\pi}{6}=\frac{3\pi-2\pi}{12}=\frac{\pi}{12}.$

   Step 4 — raise to the 50th power (De Moivre).

   $\arg\!\left(\frac{Z_1}{Z_2}\right)^{50}=50\cdot\frac{\pi}{12}=\frac{25\pi}{6}.$ …

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9. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.The domain of the real valued function $f(x) = \sqrt{\cos(\sin x)} + \cos^{-1} \left( \frac{1 + x^2}{2x} \right)$ is (A) $(-1, 1)$ (B) $[-1, 1]$ (C) $\mathbb{R} - (-1, 1)$ (D) $\{-1, 1\}$

   ›Reveal solutionSolution

   The domain is the set of real numbers where both terms are defined: the square root requires $\cos(\sin x) \ge 0$ (true for all real $x$), and the inverse cosine requires $\left|\frac{1+x^2}{2x}\right| \le 1$ and $x \neq 0$, which forces $x = \pm 1$. So the domain is $\{-1, 1\}$, option (D).

   ---

   The key idea: a function defined by a sum is only valid where every piece is defined. Here we have a square root and an inverse cosine. The square root’s domain is where its inside is non‑negative; the inverse cosine’s domain is where its argument lies in $[-1, 1]$. We find the intersection of these conditions.

   Why this approach works:

   Instead of guessing, we treat each term separately. The square root term looks scary but simplifies because $\sin x$ is bounded and cosine is always non‑negative on the range of $\sin x$. The inverse cosine term is the restrictive one — it forces a very narrow set of $x$ values.

   ---

   1. First term: $\sqrt{\cos(\sin x)}$

      For the square root to be real, we need $\cos(\sin x) \ge 0$.

      Since $\sin x \in [-1, 1]$, we check: for any $y \in [-1, 1]$, is $\cos y \ge 0$?

      The cosine function is non‑negative on $[-\pi/2, \pi/2]$, and $[-1, 1]$ is a subset of that interval (because $1 < \pi/2 \approx 1.57$).

      Therefore $\cos(\sin x) > 0$ for all real $x$.

      So the first term imposes no restriction — its domain is $\mathbb{R}$.

   2. Second term: $\cos^{-1}\left( \frac{1 + x^2}{2x} \right)$

      The inverse cosine function $\cos^{-1}(t)$ is defined only for $t \in [-1, 1]$.

      So we require

   $-1 \le \frac{1 + x^2}{2x} \le 1.$

   Also, the denominator $2x$ cannot be zero, so $x \neq 0$.

   3. Solve the inequality Consider the right inequality first:

   $\frac{1 + x^2}{2x} \le 1.$

   Bring to one side:

   $\frac{1 + x^2}{2x} - 1 \le 0 \quad\Rightarrow\quad \frac{1 + x^2 - 2x}{2x} \le 0 \quad\Rightarrow\quad \frac{(x-1)^2}{2x} \le 0.$

   The numerator $(x-1)^2$ is always $\ge 0$ and equals $0$ only at $x=1$.

   So the fraction is $\le 0$ only when the denominator is negative (since numerator is non‑negative).

   That means $2x < 0$, i.e. $x < 0$, or numerator = 0 (i.e. $x=1$).

   So from this inequality: $x < 0$ or $x = 1$.

   Now the left inequality:

   $-1 \le \frac{1 + x^2}{2x} \quad\Rightarrow\quad \frac{1 + x^2}{2x} + 1 \ge 0 \quad\Rightarrow\quad \frac{1 + x^2 + 2x}{2x} \ge 0 \quad\Rightarrow\quad \frac{(x+1)^2}{2x} \ge 0.$

   The numerator $(x+1)^2 \ge 0$, zero only at $x=-1$. …

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10. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
    Q.If $z = (\sqrt{3} - i)^{2025} + (-1 - \sqrt{3}i)^{2026}$ then the point corresponding to $z$ in Argand plane lies in (A) $1^{\text{st}}$ quadrant (B) $2^{\text{nd}}$ quadrant (C) $3^{\text{rd}}$ quadrant (D) $4^{\text{th}}$ quadrant

    ›Reveal solutionSolution

    $z = -2^{2025} + i\,2^{2025}(1-\sqrt3)$: real part $<0$ and imaginary part $<0$, so $z$ lies in the third quadrant.

    First term $(\sqrt3 - i)^{2025}$: here $\sqrt3 - i = 2\big(\cos(-\tfrac{\pi}{6}) + i\sin(-\tfrac{\pi}{6})\big)$, so

    $(\sqrt3 - i)^{2025} = 2^{2025}\Big(\cos\tfrac{-2025\pi}{6} + i\sin\tfrac{-2025\pi}{6}\Big).$

    Reducing the angle: $-\tfrac{2025\pi}{6} = -337.5\pi \equiv \tfrac{\pi}{2} \pmod{2\pi}$, so this term $= 2^{2025}\,i$.

    Second term $(-1 - \sqrt3\,i)^{2026}$: here $-1-\sqrt3\,i = 2\big(\cos\tfrac{4\pi}{3} + i\sin\tfrac{4\pi}{3}\big)$, so

    $(-1 - \sqrt3\,i)^{2026} = 2^{2026}\Big(\cos\tfrac{2026\cdot 4\pi}{3} + i\sin\tfrac{2026\cdot 4\pi}{3}\Big).$

    Reducing: $\tfrac{2026\cdot 4\pi}{3} = \tfrac{8104\pi}{3} \equiv \tfrac{4\pi}{3}\pmod{2\pi}$, giving …

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11. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
    Q.$x$ and $y$ are two complex numbers such that $|x| = |y| = 1$. If $\operatorname{Arg}(x) = 2\alpha$, $\operatorname{Arg}(y) = 3\beta$ and $\alpha + \beta = \dfrac{\pi}{36}$, then $x^6 y^4 + \dfrac{1}{x^6 y^4} =$ (A) $0$ (B) $-1$ (C) $1$ (D) $\dfrac{1}{2}$

    ›Reveal solutionSolution

    The key is to express $x^6 y^4$ in polar form using the given arguments, simplify the exponent using $\alpha+\beta = \pi/36$, and then evaluate $z + 1/z$ for a complex number on the unit circle — the result is $1$.

    We are given two complex numbers $x$ and $y$ on the unit circle ($|x|=|y|=1$), with arguments $\operatorname{Arg}(x)=2\alpha$ and $\operatorname{Arg}(y)=3\beta$, and the condition $\alpha+\beta = \pi/36$.

    We need $x^6 y^4 + \frac{1}{x^6 y^4}$.

    Concept & Intuition

    When a complex number lies on the unit circle, its reciprocal equals its conjugate. So $z + 1/z = z + \overline{z} = 2\operatorname{Re}(z)$.

    Thus the problem reduces to finding the real part of $x^6 y^4$. Since $x$ and $y$ are on the unit circle, their powers are also on the unit circle, and their arguments add. The given relation $\alpha+\beta = \pi/36$ will simplify the total argument to a nice angle whose cosine we know exactly.

    Step-by-step

    1. Write $x$ and $y$ in polar form Since $|x|=|y|=1$, we have

    $x = e^{i\cdot 2\alpha}, \qquad y = e^{i\cdot 3\beta}.$

    2. Compute $x^6 y^4$

    $x^6 = (e^{i\cdot 2\alpha})^6 = e^{i\cdot 12\alpha}, \qquad y^4 = (e^{i\cdot 3\beta})^4 = e^{i\cdot 12\beta}.$

    Hence

    $x^6 y^4 = e^{i(12\alpha + 12\beta)} = e^{i\cdot 12(\alpha+\beta)}.$

    3. Use the given $\alpha+\beta = \pi/36$ $$$$

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12. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
    Q.Let $z = x + iy$ and $P(x, y)$ be a point on the Argand plane. If $z$ satisfies the condition

    [!FORMULA] $\operatorname{Arg} \left( \frac{z - 3i}{z + 2i} \right) = \frac{\pi}{4}$

    then the locus of $P$ is (A) $x^2 + y^2 - y - 6 = 0, (x, y) \neq (0, -2)$ (B) $x^2 + y^2 - x - y - 6 = 0, (x, y) \neq (0, -2)$ (C) $x^2 + y^2 + 5x - y - 6 = 0, (x, y) \neq (0, -2)$ (D) $x^2 + y^2 + x - y - 6 = 0, (x, y) \neq (0, -2)$

    ›Reveal solutionSolution

    Rationalising $\dfrac{z-3i}{z+2i}$ and setting its argument to $\tfrac{\pi}{4}$ gives $x^2+y^2+5x-y-6=0$ (with $(0,-2)$ excluded) — option (C).

    Let $z = x+iy$, so $z-3i = x+i(y-3)$ and $z+2i = x+i(y+2)$. Multiply by the conjugate of the denominator:

    $\frac{z-3i}{z+2i} = \frac{[x+i(y-3)][x-i(y+2)]}{x^2+(y+2)^2}.$

    Expand the numerator:

    • Real part: $x^2 + (y-3)(y+2) = x^2 + y^2 - y - 6.$
    • Imaginary part: $x(y-3) - x(y+2) = -5x.$

    So the number is

    $\frac{(x^2+y^2-y-6) - 5x\,i}{x^2+(y+2)^2}.$

    Since $\operatorname{Arg} = \dfrac{\pi}{4}$ and $\tan\dfrac{\pi}{4}=1$, the imaginary and real parts are equal (both positive): …

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