Complexity! (1)

Algebra Level 3

{ z = x − i y z 3 = p + i q \begin{cases} \ \ \ z=x-iy \\ \sqrt[3]z=p+iq \end{cases}

Complex number z z satisfies the system of equations above. Find the value of ( x p + y q ) p 2 + q 2 \dfrac{\left( \frac{x}{p} + \frac{y}{q} \right) }{p^2 + q^2} .

Notation: i = − 1 i = \sqrt{-1} denotes the imaginary unit .

For more problems on complex numbers, click here .


The answer is -2.

Tapas Mazumdar by Tapas Mazumdar , 20, India

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2 solutions

Chew-Seong Cheong
Sep 23, 2016

z 3 = p + i q z = ( p + i q ) 3 x − i y = p 3 + i 3 p 2 q − 3 p q 2 − i q 3 \begin{aligned} \sqrt[3] z & = p + iq \\ z & = (p + iq)^3 \\ x - iy & = p^3 + i3p^2q - 3pq^2 - iq^3 \end{aligned}

Equating the real and imaginary parts on both sides,

{ x = p 3 − 3 p q 2 ⟹ x p = p 2 − 3 q 2 y = q 3 − 3 p 2 q ⟹ y q = q 2 − 3 p 2 \begin{cases} x = p^3 - 3pq^2 & \implies \dfrac xp = p^2 - 3q^2 \\ y = q^3 - 3p^2q & \implies \dfrac yq = q^2 - 3p^2 \end{cases}

Therefore, we have:

x p + y q p 2 + q 2 = p 2 − 3 q 2 + q 2 − 3 p 2 p 2 + q 2 = − 2 p 2 − 2 q 2 p 2 + q 2 = − 2 \begin{aligned} \frac {\frac xp + \frac yq}{p^2+q^2} & = \frac {p^2 - 3q^2 + q^2 - 3p^2}{p^2+q^2} = \frac {-2p^2-2q^2}{p^2+q^2} = \boxed{-2} \end{aligned}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Just to find answer, consider 1 and its complex cube root w.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Yeah in the exam would have done that!

Prakhar Bindal - 4 years, 8 months ago

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