Question

If $z=x-iy$ and $z^{1/3}=p+iq$, then $\frac{1}{p^{2}+q^{2}}\left(\frac{x}{p}+\frac{y}{q}\right)$ is equal to

• A. -2
• B. -1
• C. 1
• D. 2
• E. 0

Hint:

Algebra Tip: Pay close attention to negative signs when equating imaginary parts! Since the problem gave $z = x - iy$ rather than $x + iy$, equating the imaginary part correctly to $-y$ is crucial to avoiding sign errors.

Answer 1

The Correct Option is A

Concept:
When dealing with fractional powers of complex numbers, it is often easier to isolate the linear complex variable by raising both sides to the reciprocal power. After expanding, you can equate the real parts and imaginary parts respectively.

Step 1: Isolate z by cubing both sides.
Given $z^{1/3} = p + iq$, cube both sides to express $z$ in terms of $p$ and $q$: $$z = (p + iq)^3$$

Step 2: Expand the cubic expression.
Use the binomial expansion $(A+B)^3 = A^3 + 3A^2B + 3AB^2 + B^3$: $$z = p^3 + 3p^2(iq) + 3p(iq)^2 + (iq)^3$$ Since $i^2 = -1$ and $i^3 = -i$: $$z = p^3 + 3p^2qi - 3pq^2 - q^3i$$

Step 3: Equate real and imaginary parts.
Group the real terms and imaginary terms: $$z = (p^3 - 3pq^2) + i(3p^2q - q^3)$$ We are given $z = x - iy$. Therefore: Real part: $x = p^3 - 3pq^2$ Imaginary part: $-y = 3p^2q - q^3 \implies y = q^3 - 3p^2q$

Step 4: Find expressions for x/p and y/q.
Divide the $x$ equation by $p$, and the $y$ equation by $q$: $$\frac{x}{p} = p^2 - 3q^2$$ $$\frac{y}{q} = q^2 - 3p^2$$ Now, add them together: $$\frac{x}{p} + \frac{y}{q} = (p^2 - 3q^2) + (q^2 - 3p^2) = -2p^2 - 2q^2 = -2(p^2 + q^2)$$

Step 5: Substitute into the final target expression.
Substitute the result from Step 4 into the expression we need to evaluate: $$\frac{1}{p^2+q^2} \left( \frac{x}{p} + \frac{y}{q} \right) = \frac{1}{p^2+q^2} \left( -2(p^2 + q^2) \right)$$ The $(p^2+q^2)$ terms cancel out: $$= -2$$ Hence the correct answer is (A) -2.