Question

If \(z=\frac{1}{2}(i \sqrt{3}-1)\), then the value of \(\left(z-z^{2}+\right.\) \(\left.2 z^{2}\right)\left(2-z+z^{2}\right)\) is (1) 3 (2) 7 (3) \(-1\) (4) 5

Short answer

-2

Step-by-step solution

Simplify the given complex number

First, write the provided complex number in standard form. Given: \[ z = \frac{1}{2}(i \sqrt{3} - 1) \]Simplify: \[ z = \frac{1}{2}i \sqrt{3} - \frac{1}{2} \]So, \[ z = -\frac{1}{2} + \frac{\sqrt{3}}{2}i \]

Calculate the square of z

Now, let's find \(z^2\): \[ z = -\frac{1}{2} + \frac{\sqrt{3}}{2}i \]\[ z^2 = \left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)^2 \]\[ z^2 = \left(-\frac{1}{2}\right)^2 + 2\left(-\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}i\right) + \left(\frac{\sqrt{3}}{2}i\right)^2 \]\[ z^2 = \frac{1}{4} - \frac{\sqrt{3}}{2}i + \frac{3}{4}(-1) \]\[ z^2 = \frac{1}{4} - \frac{3}{4} - \frac{\sqrt{3}}{2}i \]\[ z^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i \]

Express and compute \[z - z^2\]

Calculate \(z - z^2\): \[ z - z^2 = \left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) - \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) \]Combine like terms: \[ z - z^2 = \frac{\sqrt{3}}{2}i + \frac{\sqrt{3}}{2}i \]\[ z - z^2 = \sqrt{3}i \]

Compute \ 2z^2 \

Now, calculate \[ 2z^2 \]\[ 2z^2 = 2\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) \]\[ 2z^2 = -1 - \sqrt{3}i \]

Combine the results

Combine the results to evaluate the final expression: \[ \left( z - z^2 + 2z^2 \right)\left(2 - z + z^2\right) \]Plug in the simplified terms: \[ \left( \sqrt{3}i - 1 - \sqrt{3}i \right)\left( 2 - \left( -\frac{1}{2} + \frac{\sqrt{3}}{2}i \right) + \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i \right) \]Simplify within the parentheses: \[ \left(-1\right) \left( 2 + \frac{1}{2} - \frac{1}{2} \right) \left( -1 \right) \left( 2 \right) \]Combine results: \[ = -1 \cdot 2 \]\[= -1 \cdot 2 = -2 \]

Key concepts

complex numbers

Complex numbers are a combination of a real part and an imaginary part and are written in the form of \(a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit with the property \(i^2 = -1\). For instance, in the given problem, the complex number is given as \[ z = \frac{1}{2}(i \sqrt{3} - 1) \]. To simplify it into standard form, we need to distribute the fraction: \[\begin{equation} z = \frac{1}{2}i \sqrt{3} - \frac{1}{2} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i \end{equation}\]
Understanding how to break down and convert complex numbers into this form is crucial, especially when preparing for exams like IIT JEE.

algebraic manipulation

Algebraic manipulation involves rearranging and simplifying expressions to solve for a desired variable or to make complex equations more manageable. This lesson is divided into a few steps:

• First, we simplified the given complex number into its standard form: \[\begin{equation} z = -\frac{1}{2} + \frac{\sqrt{3}}{2}i \end{equation}\]
• Next, we calculated the square of this complex number: \[\begin{equation} z^2 = \left( -\frac{1}{2} + \frac{\sqrt{3}}{2}i \right)^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i \end{equation}\]
• Then, we found the difference between \[\begin{equation} z \end{equation}\] and \[\begin{equation} z^2 \end{equation}\], and simplified to get \[ z - z^2 = \sqrt{3}i \]
• Finally, we calculated \[ 2z^2 = -1 - \sqrt{3}i \] and combined all these results to evaluate the complex expression.

This strategic breakdown highlights how algebraic manipulation is essential to solving complex number problems in exams like IIT JEE.

IIT JEE preparation

IIT JEE is one of the toughest entrance exams, requiring deep understanding and practice in various topics including complex numbers. Mastering complex numbers for IIT JEE includes:

• Learning to express complex numbers in standard form \(a + bi\)
• Understanding how to perform basic operations like addition, subtraction, multiplication, and division on complex numbers
• Practicing how to find the modulus and argument of a complex number
• Becoming proficient in operations such as calculating the square of a complex number and manipulating the expressions algebraically
• Solving past year IIT JEE problems and taking mock exams to get familiar with the pattern and complexity

By consistently practicing these skills, students can handle complex number problems effectively during their IIT JEE exam.

square of complex number

Finding the square of a complex number involves expanding the expression accurately. To find \[\begin{equation} z^2 \end{equation}\], if \[\begin{equation} z = -\frac{1}{2} + \frac{\sqrt{3}}{2}i \end{equation}\], we use the formula:
\[ z^2 = \left( -\frac{1}{2} + \frac{\sqrt{3}}{2}i \right)^2 = \left( -\frac{1}{2} \right)^2 + 2 \left( -\frac{1}{2} \right) \left( \frac{\sqrt{3}}{2}i \right) + \left( \frac{\sqrt{3}}{2}i \right)^2 \]
Calculating each part individually:
\[ \( -\frac{1}{2} \)^2 = \frac{1}{4}, \ 2 \( -\frac{1}{2} \) \( \frac{\sqrt{3}}{2}i \) = -\frac{\sqrt{3}}{2}i, \ \( \frac{\sqrt{3}}{2}i \)^2 = \frac{3}{4} \( -1 \) = -\frac{3}{4} \]
Summing these components gives:
\[ z^2 = \frac{1}{4} - \frac{3}{4} - \frac{\sqrt{3}}{2}i = -\frac{1}{2} - \frac{\sqrt{3}}{2}i \]
Carefully performing each step helps avoid mistakes, ensuring accuracy in results, which is particularly important for high-stakes exams like the IIT JEE.