Let z₁ and z₂ Be Two Complex Numbers Such That z₁ + z₂ = 5 and z₁³ + z₂³ = 20 + 15i. Then |z₁⁴ + z₂⁴| Equals

JEE Main 2024 — 31 January, Shift 2. Previous Year Question.

Problem

Let $z_1$ and $z_2$ be two complex numbers such that $z_1+z_2=5$ and $z_1^3+z_2^3=20+15i$. Then $|z_1^4+z_2^4|$ equals:

Key insight. Everything here is built from ordinary algebraic identities applied to $z_1+z_2$ and $z_1z_2$ — no separate values of $z_1$ and $z_2$ are ever needed, since every power sum asked for can be built up step by step from these two quantities alone.

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Approach

Use the identity $a^3+b^3=(a+b)^3-3ab(a+b)$ to extract $z_1z_2$ from the given cube-sum equation. With both $z_1+z_2$ and $z_1z_2$ known, build up to $z_1^2+z_2^2$ using $a^2+b^2=(a+b)^2-2ab$, and then to $z_1^4+z_2^4$ using the same squaring idea applied one level higher: $a^4+b^4=(a^2+b^2)^2-2(ab)^2$. Finally, take the modulus of the resulting complex number.

Solution

Step 1 — Extract z1z2 from the cube-sum identity

$$z_1^3+z_2^3 = (z_1+z_2)^3-3z_1z_2(z_1+z_2)$$

With $z_1+z_2=5$, so $(z_1+z_2)^3=125$:

$$20+15i = 125-15z_1z_2$$

$$15z_1z_2 = 125-(20+15i) = 105-15i$$

$$z_1z_2 = \frac{105-15i}{15} = 7-i$$

Step 2 — Find z1² + z2²

$$z_1^2+z_2^2 = (z_1+z_2)^2-2z_1z_2 = 5^2-2(7-i) = 25-14+2i = 11+2i$$

Step 3 — Find (z1z2)²

$$(z_1z_2)^2 = (7-i)^2 = 49-14i+i^2 = 49-14i-1 = 48-14i$$

Step 4 — Find z1⁴ + z2⁴

$$z_1^4+z_2^4 = (z_1^2+z_2^2)^2-2(z_1z_2)^2$$

First, $(z_1^2+z_2^2)^2$:

$$(11+2i)^2 = 121+44i+4i^2 = 121+44i-4 = 117+44i$$

Then $2(z_1z_2)^2$:

$$2(48-14i) = 96-28i$$

Subtracting:

$$z_1^4+z_2^4 = (117+44i)-(96-28i) = 21+72i$$

Step 5 — Take the modulus

$$|z_1^4+z_2^4| = \sqrt{21^2+72^2} = \sqrt{441+5184} = \sqrt{5625} = 75$$

Answer

Common mistakes

• Trying to solve for $z_1$ and $z_2$ individually. Since only $z_1+z_2$ and (eventually) $z_1z_2$ are given or derivable, solving the underlying quadratic for the actual complex roots is unnecessary extra work — every quantity asked for can be built from the sum and product alone.
• Sign errors when expanding $(11+2i)^2$ or $(7-i)^2$. The $i^2=-1$ step is where a dropped negative sign most often creeps in — it’s worth expanding each square carefully term by term rather than trying to do it in one line.
• Forgetting the factor of 2 in $a^4+b^4=(a^2+b^2)^2-2(ab)^2$. This identity is one level removed from the more familiar $a^2+b^2=(a+b)^2-2ab$, and it’s easy to drop the “$-2(ab)^2$” correction entirely and just report $(z_1^2+z_2^2)^2$.

Practise next

• If $z_1+z_2=4$ and $z_1^3+z_2^3=28+96i$, find $|z_1^4+z_2^4|$ using the same chain of identities.