Roots of 2z²−3z−2i=0 — Find 16·Re(E)·Im(E) for a Power-Sum Ratio

JEE Main 2025 — 24 January, Morning Shift. Previous Year Question.

Problem

If $\alpha$ and $\beta$ are the roots of the equation $2z^2-3z-2i=0$, where $i=\sqrt{-1}$, then

$$16\cdot\text{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)\cdot\text{Im}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)$$

is equal to:

Key insight. Directly computing $\alpha^{19}$ or $\beta^{19}$ is out of the question — but a general identity connects power sums at different exponents: $S_nS_k = S_{n+k}+(\alpha\beta)^kS_{n-k}$, where $S_n=\alpha^n+\beta^n$. Applying this with $n=15$, $k=4$ links $S_{19}$, $S_{15}$, and $S_{11}$ together — and since $\alpha\beta=-i$ raises to exactly $1$ when taken to the 4th power, the whole messy ratio collapses to a single, much smaller power sum: $S_4$.

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Approach

Find $\alpha+\beta$ and $\alpha\beta$ from Vieta’s formulas. Use the power-sum identity $S_nS_k=S_{n+k}+(\alpha\beta)^kS_{n-k}$ with $n=15,k=4$ to express $S_{19}$ in terms of $S_{15}$, $S_4$, and $S_{11}$. Substitute into the given ratio, and use the fact that $(\alpha\beta)^4=1$ to collapse the expression to just $S_4$, which can then be computed directly using the standard power-sum recursion.

Solution

Step 1 — Find α + β and αβ using Vieta’s formulas

For $2z^2-3z-2i=0$: $\alpha+\beta=\dfrac{3}{2}$, $\alpha\beta = \dfrac{-2i}{2} = -i$.

Step 2 — State the power-sum identity

For $S_n=\alpha^n+\beta^n$, using $q=\alpha\beta$:

$$S_n\cdot S_k = S_{n+k}+q^k\cdot S_{n-k}$$

With $n=15$, $k=4$: $S_{15}S_4 = S_{19}+q^4S_{11}$, so:

$$S_{19} = S_{15}S_4 – q^4S_{11}$$

Step 3 — Substitute into the given ratio

$$E = \frac{S_{19}+S_{11}}{S_{15}} = \frac{S_{15}S_4-q^4S_{11}+S_{11}}{S_{15}} = S_4 + \frac{S_{11}(1-q^4)}{S_{15}}$$

Step 4 — Evaluate q⁴

$$q^4 = (\alpha\beta)^4 = (-i)^4 = \big[(-i)^2\big]^2 = (-1)^2 = 1$$

Step 5 — Simplify E

Since $q^4=1$, the fraction $\dfrac{S_{11}(1-q^4)}{S_{15}}=0$:

$$E = S_4$$

Step 6 — Compute S₁, S₂, S₄ using the standard recursion

$$S_1 = \alpha+\beta = \frac{3}{2}$$

$$S_2 = S_1^2-2q = \left(\frac{3}{2}\right)^2-2(-i) = \frac{9}{4}+2i$$

$$S_4 = S_2^2-2q^2$$

First, $S_2^2 = \left(\frac{9}{4}+2i\right)^2 = \frac{81}{16}+2\cdot\frac{9}{4}\cdot2i+(2i)^2 = \frac{81}{16}+9i-4 = \frac{17}{16}+9i$

And $q^2=(-i)^2=-1$, so $-2q^2=2$:

$$S_4 = \frac{17}{16}+9i+2 = \frac{49}{16}+9i$$

Step 7 — Extract Re(E) and Im(E)

$$\text{Re}(E) = \frac{49}{16}, \qquad \text{Im}(E) = 9$$

Step 8 — Compute the final answer

$$16\cdot\text{Re}(E)\cdot\text{Im}(E) = 16\times\frac{49}{16}\times9 = 49\times9 = 441$$

Answer

Common mistakes

• Trying to compute $S_{19}$ and $S_{11}$ directly via the basic recursion $S_n=(\alpha+\beta)S_{n-1}-\alpha\beta S_{n-2}$. This works in principle but requires stepping through 15+ iterations — the power-sum identity linking $S_{19}$, $S_{15}$, $S_{11}$ directly is far faster once the exponents are noticed to be equally spaced ($19=15+4$, $11=15-4$).
• Forgetting to check whether $q^4=1$ before simplifying. This is the single fact that makes the entire problem tractable — without recognising $(-i)^4=1$, the term $\frac{S_{11}(1-q^4)}{S_{15}}$ wouldn’t vanish, and the ratio wouldn’t reduce to a simple power sum.

Practise next

• If $\gamma,\delta$ are the roots of $z^2-z-i=0$, find $16\cdot\text{Re}(F)\cdot\text{Im}(F)$ where $F=\dfrac{\gamma^{13}+\delta^{13}+\gamma^{5}+\delta^{5}}{\gamma^{9}+\delta^{9}}$, using the same power-sum identity approach.