What is the conjugate of $\dfrac{2-i}{(1-2i)^2}$?

> [!TLDR] > To find the conjugate of a complex fraction, first simplify the fraction into the standard $a+bi$ form by expanding the denominator and then rationalizing. Once in $a+bi$ form, the conjugate is $a-bi$. The conjugate of $\dfrac{2-i}{(1-2i)^2}$ is $\boxed{-\frac{2}{25} - \frac{11}{25}i}$. When working with complex numbers, the concept of a conjugate is fundamental. For any complex number $z = a+bi$, its conjugate, denoted $\overline{z}$, is $a-bi$. Geometrically, this is a reflection across the real axis in the complex plane. The key to solving this problem lies in understanding how conjugates interact with arithmetic operations. A very useful property is that the conjugate of a sum, difference, product, or quotient is the sum, difference, product, or quotient of the conjugates, respectively. That is: * $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$ * $\overline{z_1 - z_2} = \overline{z_1} - \overline{z_2}$ * $\overline{z_1 z_2} = \overline{z_1} \overline{z_2}$ * $\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\overline{z_1}}{\overline{z_2}}$ (provided $z_2 \neq 0$) * $\overline{z^n} = (\overline{z})^n$ These properties mean we have two main approaches: 1. Simplify the entire expression into the form $a+bi$ first, and then take its conjugate. 2. Take the conjugate of each part of the expression first, and then simplify. For this particular problem, simplifying the expression first is generally more straightforward. We will first transform the given complex fraction into the standard $a+bi$ form, and then apply the definition of the conjugate. 1. **Simplify the denominator:** The denominator is $(1-2i)^2$. We expand this using the formula $(x-y)^2 = x^2 - 2xy + y^2$: $$ (1-2i)^2 = 1^2 - 2(1)(2i) + (2i)^2 $$ Recall that $i^2 = -1$. $$ (1-2i)^2 = 1 - 4i + 4i^2 $$ $$ (1-2i)^2 = 1 - 4i + 4(-1) $$ $$ (1-2i)^2 = 1 - 4i - 4 $$ $$ (1-2i)^2 = -3 - 4i $$ So, the original expression becomes $\dfrac{2-i}{-3-4

What is the conjugate of (2-i)/((1-2i)²)?

NCERT Exemplar · Q11

\dfrac{2-i}{(1-2i)^2}

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Concept understanding — Complex Number Arithmetic

Complex Number Arithmetic: A First Look

Imagine you're trying to solve $x^2 + 1 = 0$ . You know that no real number squared gives $-1$ . The square of any real number is either zero or positive. So this equation has no real solution. But what if we invent a number whose square is $-1$ ? That's exactly what mathematicians did — and that invention is the imaginary unit $i$ , defined by:

i^2 = -1

A complex number is any number of the form $a + bi$ , where $a$ and $b$ are real numbers. Here $a$ is called the real part, and $b$ is called the imaginary part. For example, $3 + 4i$ has real part $3$ and imaginary part $4$ .

Note

The name "imaginary" is unfortunate — these numbers are just as real (in the mathematical sense) as the numbers you already know. They're simply a different kind of number.

Why Bother?

Complex numbers let you solve equations that real numbers can't. Every polynomial equation — no matter how complicated — has a solution in the complex numbers. This is the Fundamental Theorem of Algebra, and it's one of the most important results in mathematics.

Arithmetic Operations

The rules are straightforward: treat $i$ like a variable, but remember that $i^2 = -1$ .

Addition and Subtraction

Add (or subtract) real parts with real parts, imaginary parts with imaginary parts.

(a + bi) + (c + di) = (a + c) + (b + d)i

(a + bi) - (c + di) = (a - c) + (b - d)i

Example: $(2 + 3i) + (4 - 5i) = (2 + 4) + (3 - 5)i = 6 - 2i$

Multiplication

Multiply like binomials, then replace $i^2$ with $-1$ .

(a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i

Example: $(2 + 3i)(4 - 5i) = 2(4) + 2(-5i) + 3i(4) + 3i(-5i)$

$= 8 - 10i + 12i - 15i^2$

$= 8 + 2i - 15(-1)$

$= 8 + 2i + 15 = 23 + 2i$

Watch out

The most common mistake: forgetting that $i^2 = -1$ , not $1$ . Always check your final step.

Division

Division is trickier. The key idea: multiply numerator and denominator by the complex conjugate of the denominator.

The complex conjugate of $a + bi$ is $a - bi$ . When you multiply a complex number by its conjugate, you get a real number:

(a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2 i^2 = a^2 + b^2

So to divide:

\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(a + bi)(c - di)}{c^2 + d^2}

Example: $\frac{2 + 3i}{4 - 5i} = \frac{(2 + 3i)(4 + 5i)}{(4 - 5i)(4 + 5i)} = \frac{8 + 10i + 12i + 15i^2}{16 + 25} = \frac{8 + 22i - 15}{41} = \frac{-7 + 22i}{41} = -\frac{7}{41} + \frac{22}{41}i$

Tip

To divide quickly: multiply top and bottom by the conjugate of the denominator, then simplify. The denominator always becomes $c^2 + d^2$ , a positive real number.

The Big Picture …

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To find the conjugate of a complex fraction, first simplify the fraction into the standard $a+bi$ form by expanding the denominator and then rationalizing. Once in $a+bi$ form, the conjugate is $a-bi$ . The conjugate of $\dfrac{2-i}{(1-2i)^2}$ is $\boxed{-\frac{2}{25} - \frac{11}{25}i}$ .

When working with complex numbers, the concept of a conjugate is fundamental. For any complex number $z = a+bi$ , its conjugate, denoted $\overline{z}$ , is $a-bi$ . Geometrically, this is a reflection across the real axis in the complex plane.

The key to solving this problem lies in understanding how conjugates interact with arithmetic operations. A very useful property is that the conjugate of a sum, difference, product, or quotient is the sum, difference, product, or quotient of the conjugates, respectively. That is:

$\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$
$\overline{z_1 - z_2} = \overline{z_1} - \overline{z_2}$
$\overline{z_1 z_2} = \overline{z_1} \overline{z_2}$
$\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\overline{z_1}}{\overline{z_2}}$ (provided $z_2 \neq 0$ )
$\overline{z^n} = (\overline{z})^n$

These properties mean we have two main approaches:

Simplify the entire expression into the form $a+bi$ first, and then take its conjugate.
Take the conjugate of each part of the expression first, and then simplify.

For this particular problem, simplifying the expression first is generally more straightforward. We will first transform the given complex fraction into the standard $a+bi$ form, and then apply the definition of the conjugate.

Simplify the denominator: The denominator is $(1-2i)^2$ . We expand this using the formula $(x-y)^2 = x^2 - 2xy + y^2$ :

(1-2i)^2 = 1^2 - 2(1)(2i) + (2i)^2

Recall that $i^2 = -1$.

(1-2i)^2 = 1 - 4i + 4i^2

(1-2i)^2 = 1 - 4i + 4(-1)

(1-2i)^2 = 1 - 4i - 4

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

So, the original expression becomes $\dfrac{2-i}{-3-4i}$.

2. Express the fraction in standard $a+bi$ form:

To express a complex fraction $\dfrac{z_1}{z_2}$ in the form $a+bi$ , we multiply both the numerator and the denominator by the conjugate of the denominator. The denominator is $-3-4i$ , so its conjugate is $-3+4i$ .

$\frac{2-i}{-3-4i} = \frac{2-i}{-3-4i} \times \frac{-3+4i}{-3+4i}$ …

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Q.What is the conjugate of 2−i(1−2i)2\dfrac{2-i}{(1-2i)^2}(1−2i)22−i​?

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Q.What is the conjugate of $\dfrac{2-i}{(1-2i)^2}$ ?

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