Problem 75 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(z=1+a i\) be a complex number, \(a>0\), such that \(z^{3}\) is areal number. Then the sum \(1+z+z^{2}+\ldots+z^{11}\) is equal to: (Online April 10,2016) (a) \(1365 \sqrt{3} \mathrm{i}\) (b) \(-1365 \sqrt{3} \mathrm{i}\) (c) \(-1250 \sqrt{3} \mathrm{i}\) (d) \(1250 \sqrt{3} \mathrm{i}\)

Step-by-Step Solution

Answer

The sum is \(1365 \sqrt{3} i\), option (a).

Step 1: Find the argument for z and z^3

Given that \(z = 1 + ai\) and \(z^3\) is a real number, it implies that \(z^3 = x + 0i\) for some real \(x\). For a complex number to be real, its imaginary part must be zero, which sometimes results in the angle from the positive real axis being a multiple of \(\pi\). This means the argument of \(z^3\) (denoted as \(3\theta\)) must be integer multiple of \(\pi\). Hence, \(3\theta = k\pi\), where \(k\) is an integer. Since \(z = 1 + ai\), the argument of \(z\) is \(\tan^{-1}(a)\) and \(\theta = \tan^{-1}(a)\). Therefore, \(3\tan^{-1}(a) = k\pi\).

Step 2: Solve for a condition of a

From \(3\tan^{-1}(a) = \pi\), since \(a > 0\) (a must be positive), we apply \(\tan^{-1}(a) = \frac{\pi}{3}\). Thus, \(a = \tan\left(\frac{\pi}{3}\right) = \sqrt{3}\). Therefore, \(z = 1 + \sqrt{3}i\).

Step 3: Calculate z^3 to verify it is real

Now compute \(z^3 = (1 + \sqrt{3}i)^3\) to verify it is real. Using the binomial theorem: \[z^3 = (1 + \sqrt{3}i)^3 = 1 + 3(1)(\sqrt{3}i) + 3(\sqrt{3}i)^2 + (\sqrt{3}i)^3\].Calculate each term: - \(3(1)(\sqrt{3}i) = 3\sqrt{3}i\), - \(3(\sqrt{3}i)^2 = 3(-3) = -9\), - \(\sqrt{3}i^3 = 0 - 3i\). Therefore, \[z^3 = 1 + 3\sqrt{3}i - 9 - 3i = -8\].So, \(z^3\) is indeed a real number.

Step 4: Use sum formula for geometric series

The sum of the series \(1 + z + z^2 + \ldots + z^{11}\) is a geometric series with first term \(a = 1\), common ratio \(r = z\), and 12 terms (0 through 11). The sum of a geometric series is given by: \[ S_n = \frac{a(r^n - 1)}{r - 1}\],where \(n = 12\). Substitute \(a = 1\), \(r = z\), and \(n = 12\): \[ S_{12} = \frac{1-z^{12}}{1 - z} \].Since \(z^3 = -8\), we have \(z^{12} = (z^3)^4 = (-8)^4 = 4096\). Compute: \(1 - z^{12} = 1 - 4096 = -4095\), and \(1 - z = 1 - (1 + \sqrt{3}i) = -\sqrt{3}i\).

Step 5: Simplify final sum expression

The series sum simplifies to:\[ S_{12} = \frac{-4095}{-\sqrt{3}i}\].Multiply by \(\frac{\sqrt{3} i}{\sqrt{3} i}\) to simplify: \[\frac{-4095}{-\sqrt{3}i} \times \frac{\sqrt{3} i}{\sqrt{3} i} = \frac{-4095(\sqrt{3} i)}{-3}(i^2)\ = \frac{-4095(\sqrt{3}i)(-1)}{3} = 1365 \sqrt{3} i\].Hence, \(1 + z + z^2 + \ldots + z^{11} = 1365 \sqrt{3} i\).

Step 6: Determine the correct option

From the calculation, the answer is \(1365 \sqrt{3} i\), which corresponds to option (a).

Key Concepts

Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a constant, called the common ratio. Understanding geometric series is crucial when dealing with complex numbers, especially in the context of powers. To find the sum of a finite geometric series like the one given in the exercise

1. First, identify the first term \(a\) and the common ratio \(r\).
2. Calculate the sum using the formula: \[S_n = \frac{a(r^n - 1)}{r - 1}\] where \(n\) represents the number of terms in the series.

In this particular problem, the series was formed by the complex number \(z\), highlighting how geometric series formulas are applicable to complex numbers. Knowing this, you can compute complex sequences efficiently.

Binomial Theorem

The binomial theorem is a fundamental principle in algebra that provides a formula to expand powers of sums. It plays a major role in calculating powers of complex numbers like \((1 + ai)^n\). The binomial theorem states\((a + b)^n = \sum_{k=0}^{n} \binom{n}{k}a^{n-k}b^k.\)This theorem is valuable for breaking down complex expressions such as \((1 + \sqrt{3}i)^3\) into manageable calculations involving real and imaginary components. It helps achieve simplification by dividing computations into:

• Real parts and
• Imaginary parts.

Through careful application of the binomial theorem, you can verify properties or specific outputs of complex numbers, such as showing that \(z^3\) is real.

Imaginary Unit

The imaginary unit, denoted as \(i\), is a mathematical entity used to extend the real numbers to complex numbers. This is essential because it's defined by the property:\[i^2 = -1.\] When calculating powers of \(i\), you notice a cyclical pattern:

• \(i^1 = i\)
• \(i^2 = -1\)
• \(i^3 = -i\)
• \(i^4 = 1\)

Repeating every four terms, this cycle is beneficial when simplifying powers of complex numbers that include \(i\). In the context of the exercise, knowing how to manage the imaginary unit's properties ensures correct simplification of expressions, thereby finding the correct answer efficiently.

Argument of a Complex Number

The argument of a complex number is the angle \(\theta\) in the complex plane, often measured from the positive real axis to the line representing the complex number. For a given complex number \(z = a + bi\), the argument is given by \[\theta = \tan^{-1}\left(\frac{b}{a}\right).\] Understanding the argument is pivotal when dealing with power transformations of complex numbers. It assists in identifying when the number will become a purely real entity, as seen in the exercise when determining \(z^3\). This conversion is critical when considering the multiplicity of the argument because for a number to return to being real, the argument's multiple must align properly, typically matching an integer multiple of \(\pi\). By grasping this concept, you gain the insight necessary for resolving complex number transformations effectively.