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Animated Solution for Mathematics - Complex Numbers: If $z = x - i y$ and $z^{1/3} = p + i q$ , then $(x / p + y / q) / (p^{2} + q^{2})$ is equal to

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Visualized Solution

Given Equations

Given: $z = x - i y$
Given: $z^{\frac{1}{3}} = p + i q$

Eliminate the Cube Root

To find $z$ , cube both sides of $z^{\frac{1}{3}} = p + i q$ :
$z = (p + i q)^{3}$

Recall the Binomial Expansion

Use the identity: $(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$
Here, $a = p$ and $b = i q$

Apply the Expansion Formula

Substituting $a$ and $b$ into the expansion:
$z = p^{3} + 3 p^{2} (i q) + 3 p (i q)^{2} + (i q)^{3}$

Simplify Powers of Iota

Recall: $i^{2} = - 1$ and $i^{3} = - i$
Substitute these into the expression:
$z = p^{3} + 3 p^{2} q i + 3 p (- q^{2}) + (- i q^{3})$

Group Real and Imaginary Parts

Group real and imaginary terms:
$z = (p^{3} - 3 p q^{2}) + i (3 p^{2} q - q^{3})$

Compare with the Given z

We are given: $z = x - i y$
Equating real and imaginary parts:
Real part: $x = p^{3} - 3 p q^{2}$
Imaginary part: $- y = 3 p^{2} q - q^{3}$

Solve for x/p

From $x = p^{3} - 3 p q^{2}$ :
Factor out $p$ : $x = p (p^{2} - 3 q^{2})$
Divide by $p$ : $\frac{x}{p} = p^{2} - 3 q^{2}$

Solve for y/q

From $- y = 3 p^{2} q - q^{3}$ :
Multiply by $- 1$ : $y = q^{3} - 3 p^{2} q$
Factor out $q$ : $y = q (q^{2} - 3 p^{2})$
Divide by $q$ : $\frac{y}{q} = q^{2} - 3 p^{2}$

Sum the Two Ratios

Calculate the sum $\frac{x}{p} + \frac{y}{q}$ :
$\frac{x}{p} + \frac{y}{q} = (p^{2} - 3 q^{2}) + (q^{2} - 3 p^{2})$

Combine Like Terms

Combine terms:
$\frac{x}{p} + \frac{y}{q} = - 2 p^{2} - 2 q^{2}$
Factor out $- 2$ :
$\frac{x}{p} + \frac{y}{q} = - 2 (p^{2} + q^{2})$

Final Calculation

Substitute the sum into the final expression:
$\frac{\frac{x}{p} + \frac{y}{q}}{p ^{2} + q ^{2}} = \frac{- 2 ( p ^{2} + q ^{2} )}{p ^{2} + q ^{2}}$
Result: $- 2$

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The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

We are given the complex number

z = x - i y

and the relation

z^{\frac{1}{3}} = p + i q

. Our objective is to evaluate the expression:

\frac{\frac{x}{p} + \frac{y}{q}}{p ^{2} + q ^{2}}

To begin, we eliminate the cube root by cubing both sides of the equation

z^{\frac{1}{3}} = p + i q

, which yields:

z = (p + i q)^{3}

Expanding the Complex Expression

Using the binomial expansion formula

(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

, we substitute

a = p

and

b = i q

z = p^{3} + 3 p^{2} (i q) + 3 p (i q)^{2} + (i q)^{3}

Recalling that

i^{2} = - 1

and

i^{3} = - i

, we simplify the expression:

z = p^{3} + 3 p^{2} q i - 3 p q^{2} - i q^{3}

Grouping the real and imaginary components, we obtain:

z = (p^{3} - 3 p q^{2}) + i (3 p^{2} q - q^{3})

Equating Real and Imaginary Parts

We compare this result to the given definition

z = x - i y

. By equating the real parts, we find:

x = p^{3} - 3 p q^{2}

By equating the imaginary parts, we find:

- y = 3 p^{2} q - q^{3} \Rightarrow y = q^{3} - 3 p^{2} q

Final Calculation

We now determine the individual ratios required for the final expression. From

x = p (p^{2} - 3 q^{2})

, we have:

\frac{x}{p} = p^{2} - 3 q^{2}

From

y = q (q^{2} - 3 p^{2})

, we have:

\frac{y}{q} = q^{2} - 3 p^{2}

Adding these two results together, we get:

\frac{x}{p} + \frac{y}{q} = (p^{2} - 3 q^{2}) + (q^{2} - 3 p^{2}) = - 2 p^{2} - 2 q^{2} = - 2 (p^{2} + q^{2})

Dividing this sum by the denominator

(p^{2} + q^{2})

, we arrive at the final result:

\frac{- 2 ( p ^{2} + q ^{2} )}{p ^{2} + q ^{2}} = - 2

Common Traps & FAQs

Wait, why is the imaginary part equated to -y and not y? The problem states z = x - iy.

This is a classic trap! When you expand (p + iq)^3, you get a term like i(3p^2q - q^3). So, the imaginary part of your expansion is (3p^2q - q^3). Since the problem defines z = x - iy, the imaginary part is -y. Therefore, -y = 3p^2q - q^3. If you forget the negative sign, your final answer will be wrong. Always map the real and imaginary components carefully!

Can't I just use De Moivre's Theorem to solve this?

You absolutely can! You could write z in polar form r(cos\theta + i sin\theta), but that introduces trigonometric functions and makes the algebra significantly more complex. In JEE, always look for the path of least resistance—algebraic expansion is much faster and less prone to calculation errors here.

What if p or q is zero? Doesn't the expression \frac{x}{p} + \frac{y}{q} become undefined?

That is a sharp observation! The expression \frac{x}{p} + \frac{y}{q} implies p \neq 0 and q \neq 0. In the context of this problem, we assume p and q are non-zero to make the expression defined. Don't let the potential for division by zero distract you from the algebraic identity being tested; focus on the manipulation of the variables.

Animated Solution for Mathematics - Complex Numbers: If $z = x - i y$ and $z^{1/3} = p + i q$ , then $(x / p + y / q) / (p^{2} + q^{2})$ is equal to

Select Answer:

Visualized Solution

Given Equations

Eliminate the Cube Root

Recall the Binomial Expansion

Apply the Expansion Formula

Simplify Powers of Iota

Group Real and Imaginary Parts

Compare with the Given z

Solve for x/p

Solve for y/q

Sum the Two Ratios

Combine Like Terms

Final Calculation

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

Expanding the Complex Expression

Equating Real and Imaginary Parts

Final Calculation

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z=x−iy and z1/3=p+iq, then (x/p+y/q)/(p2+q2) is equal to

Select Answer:

Visualized Solution

Given Equations

Eliminate the Cube Root

Recall the Binomial Expansion

Apply the Expansion Formula

Simplify Powers of Iota

Group Real and Imaginary Parts

Compare with the Given z

Solve for x/p

Solve for y/q

Sum the Two Ratios

Combine Like Terms

Final Calculation

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

Expanding the Complex Expression

Equating Real and Imaginary Parts

Final Calculation

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