A real value of $x$ satisfies the equation $\left(\dfrac{3-4ix}{3+4ix}\right)=\alpha-i\beta$ $(\alpha,\beta\in\mathbf{R})$ if $\alpha^2+\beta^2=$ (A) $1$ (B) $-1$ (C) $2$ (D) $-2$

A real value of x satisfies the equation ((3-4ix)/(3+4ix))=α-iβ (α,β∈mathbfR) if α²+β²= (A) 1 (B) -1 (C) 2 (D)

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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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The left side is a complex number of modulus 1 (ratio of conjugates), so its square modulus $\alpha^2 + \beta^2$ must equal $1$ .

When you divide one complex number by another, the modulus of the quotient is the quotient of the moduli. That geometric fact is the key here. Before diving into algebra, notice that $3 - 4ix$ and $3 + 4ix$ are conjugates of each other (when $x$ is real). The ratio of a complex number to its conjugate always lies on the unit circle.

Let me show you why, then verify it directly.

Why the modulus is 1

The numerator $3 - 4ix$ and denominator $3 + 4ix$ differ only in the sign of the imaginary part. For any complex number $z$ , we have $|z| = |\overline{z}|$ (a number and its conjugate have the same modulus). Therefore:

\left|\frac{3-4ix}{3+4ix}\right| = \frac{|3-4ix|}{|3+4ix|} = \frac{|3-4ix|}{|3-4ix|} = 1

Since the left side equals $\alpha - i\beta$ , we have $|\alpha - i\beta| = 1$ . The modulus of $\alpha - i\beta$ is $\sqrt{\alpha^2 + \beta^2}$ , so:

\sqrt{\alpha^2 + \beta^2} = 1 \implies \alpha^2 + \beta^2 = 1

Verification by direct computation

If you want to see this algebraically, multiply numerator and denominator by the conjugate of the denominator:

Rationalize the denominator:

\frac{3-4ix}{3+4ix} \cdot \frac{3-4ix}{3-4ix} = \frac{(3-4ix)^2}{(3+4ix)(3-4ix)}

Compute the denominator using the difference of squares:

(3+4ix)(3-4ix) = 9 - (4ix)^2 = 9 - 16i^2x^2 = 9 + 16x^2

Expand the numerator:

(3-4ix)^2 = 9 - 24ix + 16i^2x^2 = 9 - 24ix - 16x^2 = (9-16x^2) - 24ix

Combine: …

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Q.A real value of xxx satisfies the equation (3−4ix3+4ix)=α−iβ\left(\dfrac{3-4ix}{3+4ix}\right)=\alpha-i\beta(3+4ix3−4ix​)=α−iβ (α,β∈R)(\alpha,\beta\in\mathbf{R})(α,β∈R) if α2+β2=\alpha^2+\beta^2=α2+β2= (A) 111 (B) −1-1−1 (C) 222 (D) −2-2−2

Why the modulus is 1

Verification by direct computation

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Q.A real value of $x$ satisfies the equation $\left(\dfrac{3-4ix}{3+4ix}\right)=\alpha-i\beta$ $(\alpha,\beta\in\mathbf{R})$ if $\alpha^2+\beta^2=$
(A) $1$
(B) $-1$
(C) $2$
(D) $-2$

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