Let z1 and z2 be two complex number such that z1 + z2 = 5 and z1³ + z2³ = 20 + 15i. Then |z1⁴ + z2⁴|

Q: Wait, why don't we just solve for $z_1$ and $z_2$ using the quadratic equation $t^2 - (z_1+z_2)t + z_1z_2 = 0$?

You absolutely could! However, in JEE Advanced, time is your most precious currency. Solving for the roots involves taking square roots of complex numbers, which introduces polar forms and potential errors. Using symmetric identities is elegant, faster, and less prone to calculation slips. We want to work smarter, not harder.

Q: I always mess up the signs when squaring complex numbers like $(11+2i)^2$. Any tips?

The most common trap is forgetting that $i^2 = -1$. When you expand $(a+bi)^2$, it becomes $a^2 + 2abi + b^2i^2$, which is $(a^2 - b^2) + 2abi$. Always write out the $i^2$ term explicitly before simplifying to avoid that negative sign trap! It is a small step that saves you from a massive headache.

Q: Is there a general way to find $z_1^n + z_2^n$ for any $n$?

Yes! This is known as Newton's Sums. For any quadratic equation $t^2 - St + P = 0$, the sum of powers $S_n = z_1^n + z_2^n$ satisfies the recurrence relation $S_n - S S_{n-1} + P S_{n-2} = 0$. It is a powerful tool for higher powers and a great concept to master for competitive exams!

Question

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

Options:

(A) $30\sqrt{3}$

(B) 75

(C) $15\sqrt{15}$

(D) $25\sqrt{3}$

Sigma Percentile · Accepted Answer

### The Symphony of Symmetry: Unlocking Complex Powers

Welcome, future engineers. Today, we are not just solving a problem; we are embarking on a journey through the elegant architecture of complex numbers.

When you first look at a problem like this—where you are given the sum of two complex numbers and the sum of their cubes, and asked for the sum of their fourth powers—it is natural to feel a bit intimidated. You might be tempted to dive straight into finding $z_1$ and $z_2$ individually.

But pause. Take a breath. In the world of JEE Advanced, the most beautiful path is rarely the one that involves brute force. It is the path of symmetry.

### Phase 1: The Philosophy of Symmetry

Imagine you are standing before a grand, intricate machine. You have the input ($z_1 + z_2 = 5$) and a specific output ($z_1^3 + z_2^3 = 20 + 15i$).

Your goal is to find the state of the machine at the fourth power ($|z_1^4 + z_2^4|$). The secret here is that we do not need to know the individual values of $z_1$ and $z_2$.

We only need to know their sum, $S = z_1 + z_2$, and their product, $P = z_1z_2$. These two quantities are the DNA of the system. If we hold the sum and the product, we hold the power to generate any power of these numbers. This is the essence of symmetric polynomials.

### Phase 2: The Cubic Bridge

We begin our ascent. We are given $z_1^3 + z_2^3 = 20 + 15i$. We know the identity $z_1^3 + z_2^3 = (z_1 + z_2)^3 - 3z_1z_2(z_1 + z_2)$.

This identity is our bridge. It connects the world of cubes to the world of sums and products. Let us substitute what we know:

$$20 + 15i = (5)^3 - 3z_1z_2(5)$$

Look at that. We have transformed a cubic problem into a simple linear equation in terms of the product $P = z_1z_2$. Calculating $5^3$ gives us $125$.

So, our equation becomes $20 + 15i = 125 - 15P$. Rearranging this, we find $15P = 125 - 20 - 15i$, which simplifies to $15P = 105 - 15i$.

Dividing by $15$, we arrive at the golden key: $P = z_1z_2 = 7 - i$. We have successfully unlocked the product. Now, the rest is just a matter of climbing the ladder.

### Phase 3: Climbing the Ladder of Powers

We have the sum ($S=5$) and the product ($P=7-i$). We need to reach the fourth power. We cannot jump there in one leap, so we climb to the second power first.

The identity for the sum of squares is $z_1^2 + z_2^2 = (z_1 + z_2)^2 - 2z_1z_2$. Substituting our values:

$$z_1^2 + z_2^2 = (5)^2 - 2(7 - i)$$

$$z_1^2 + z_2^2 = 25 - 14 + 2i = 11 + 2i$$

See how clean that is? We are building our solution layer by layer. Now, for the final ascent to the fourth power.

We use the same logic, treating $z_1^2$ and $z_2^2$ as our new variables. The identity is $z_1^4 + z_2^4 = (z_1^2 + z_2^2)^2 - 2(z_1z_2)^2$.

First, calculate the squares:
$$(z_1^2 + z_2^2)^2 = (11 + 2i)^2 = 121 + 44i + 4i^2 = 121 - 4 + 44i = 117 + 44i$$
$$(z_1z_2)^2 = (7 - i)^2 = 49 - 14i + i^2 = 49 - 1 - 14i = 48 - 14i$$

Now, combine them into our fourth-power identity:

$$z_1^4 + z_2^4 = (117 + 44i) - 2(48 - 14i)$$

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

### Phase 4: The Final Modulus

We have arrived at the summit. The problem asks for the modulus of this result. The modulus of a complex number $x + iy$ is defined as $\sqrt{x^2 + y^2}$.

Here, our complex number is $21 + 72i$. So, we calculate:

$$|z_1^4 + z_2^4| = \sqrt{21^2 + 72^2}$$

$$|z_1^4 + z_2^4| = \sqrt{441 + 5184} = \sqrt{5625}$$

And the square root of $5625$ is exactly $75$. We have navigated the algebraic landscape, respected the symmetry, and arrived at the answer with precision.

The final answer is **75**. This, my friends, is the power of structured thinking. You did not just solve a problem; you mastered a technique. Keep this mindset, and no problem will ever be too complex for you.

Animated Solution for Mathematics - Complex Numbers: Let $z_{1}$ and $z_{2}$ be two complex number such that $z_{1} + z_{2} = 5$ and $z_{1}^{3} + z_{2}^{3} = 20 + 15 i$ . Then $∣ z_{1}^{4} + z_{2}^{4} ∣$ equals

Select Answer:

Visualized Solution

Analyze the Given Information

The Cubic Identity

Substitute Known Values

Solve for $z_{1} z_{2}$

Finding the Sum of Squares

Compute $z_{1}^{2} + z_{2}^{2}$

Moving to the Fourth Power

Squaring the Terms

Calculate $z_{1}^{4} + z_{2}^{4}$

Final Modulus Calculation

Conclusion

The Sigma Insight: Algebraic Operations on Complex Numbers

The Symphony of Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Let $z_{1}$ and $z_{2}$ be any two non-zero complex numbers such that $3∣ z_{1} ∣ = 4∣ z_{2} ∣$ . If $z = \frac{3 z _{1}}{2 z _{2}} + \frac{2 z _{2}}{3 z _{1}}$ then :

Let $z$ be a complex number such that $∣ z ∣ + z = 3 + i$ (where $i = - 1$ ). Then $∣ z ∣$ is equal to:

For any two complex numbers $z_{1}, z_{2}$ and any real number $a$ and $b$ , $∣ a z_{1} - b z_{2} ∣^{2} + ∣ b z_{1} + a z_{2} ∣^{2} = \dots$

If $z = 2 + 3 i$ , then $z^{5} + (\overset{z}{ˉ})^{5}$ is equal to :

Let $z$ be a complex number with non-zero imaginary part. If $\frac{2 + 3 z + 4 z ^{2}}{2 - 3 z + 4 z ^{2}}$ is a real number, then the value of $∣ z ∣^{2}$ is _____________.

Sum of squares of modulus of all the complex numbers $z$ satisfying $\overset{z}{ˉ} = i z^{2} + z^{2} - z$ is equal to \_\_\_\_\_.

Let $S = {z \in C : \overset{z}{ˉ} = i (z^{2} + Re (\overset{z}{ˉ}))}$ . Then $\sum_{z \in S} ∣ z ∣^{2}$ is equal to

If $z = \frac{1}{2} - 2 i$ , is such that $∣ z + 1∣ = α z + β (1 + i), i = - 1$ and $α, β \in R$ , then $α + β$ is equal to

Let $ω = e^{iπ /3}$ , and $a, b, c, x, y, z$ be non-zero complex numbers such that $a + b + c = x, a + bω + c ω^{2} = y, a + b ω^{2} + c ω = z$ . Then the value of $\frac{∣ x ∣ ^{2} + ∣ y ∣ ^{2} + ∣ z ∣ ^{2}}{∣ a ∣ ^{2} + ∣ b ∣ ^{2} + ∣ c ∣ ^{2}}$ is .........