Complex Numbers
Basic Complex Number Operations
Grade 11

Question:

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

Step-by-Step Solution

Key Concept: If z^(1/3) = p + iq, then z = (p + iq)³. Expand this and equate real and imaginary parts with z = x - iy to find relationships between x, y, p, q. The expression simplifies using these relationships and properties of complex cube roots.
Step 1: Express $z$ in terms of $p$ and $q$. Given $z^{1/3} = p + iq$, we cube both sides of the equation to find an expression for $z$. $$z = (p + iq)^3$$ Step 2: Expand the cubic expression. We expand the term $(p + iq)^3$ using the binomial expansion formula $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. $$(p + iq)^3 = p^3 + 3p^2(iq) + 3p(iq)^2 + (iq)^3$$ $$ = p^3 + 3ip^2q + 3p(-q^2) + i^3q^3$$ Since $i^2 = -1$ and $i^3 = -i$, we substitute these values: $$ = p^3 + 3ip^2q - 3pq^2 - iq^3$$ Step 3: Group the real and imaginary parts. Combine the real terms and the imaginary terms from the expanded expression for $z$. $$z = (p^3 - 3pq^2) + i(3p^2q - q^3)$$ Step 4: Equate real and imaginary parts to find $x$ and $y$. We are given $z = x - iy$. Comparing this with the expression for $z$ from Step 3, we equate the real parts and the imaginary parts. Equating the real parts: $$x = p^3 - 3pq^2$$ Equating the imaginary parts: $$-y = 3p^2q - q^3$$ From the second equation, we can find the expression for $y$: $$y = -(3p^2q - q^3) = q^3 - 3p^2q$$ Step 5: Calculate the value of $\left(\dfrac{x}{p} + \dfrac{y}{q}\right)$. Substitute the expressions for $x$ and $y$ obtained in Step 4 into the term $\left(\dfrac{x}{p} + \dfrac{y}{q}\right)$. First, calculate $\dfrac{x}{p}$: $$\frac{x}{p} = \frac{p^3 - 3pq^2}{p} = \frac{p(p^2 - 3q^2)}{p} = p^2 - 3q^2$$ Next, calculate $\dfrac{y}{q}$: $$\frac{y}{q} = \frac{q^3 - 3p^2q}{q} = \frac{q(q^2 - 3p^2)}{q} = q^2 - 3p^2$$ Now, sum these two expressions: $$\frac{x}{p} + \frac{y}{q} = (p^2 - 3q^2) + (q^2 - 3p^2)$$ $$ = p^2 - 3p^2 - 3q^2 + q^2$$ $$ = -2p^2 - 2q^2$$ Factor out $-2$: $$ = -2(p^2 + q^2)$$ Step 6: Calculate the final expression. Substitute the result from Step 5 into the given expression $\dfrac{\left(\dfrac{x}{p} + \dfrac{y}{q}\right)}{(p^2 + q^2)}$. $$\dfrac{\left(\dfrac{x}{p} + \dfrac{y}{q}\right)}{(p^2 + q^2)} = \dfrac{-2(p^2 + q^2)}{(p^2 + q^2)}$$ Assuming $p^2 + q^2 \neq 0$, we can cancel the term $(p^2 + q^2)$ from the numerator and the denominator. $$ = -2$$ Step 7: State the final answer. The value of the given expression is $-2$. This matches Option 2.
Correct Answer: C

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