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Algebra Level 1

The point ( 4 + 7 3 , 7 − 4 3 ) \left(4+7\sqrt{3}\ ,\ 7-4\sqrt{3}\right) is rotated π 3 \dfrac{\pi}{3} radians counterclockwise about the origin. If the resulting image is ( a , b ) (a,b) , then what is a + b a+b ?


The answer is 22.

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Andy Hayes by Andy Hayes , 38, USA

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2 solutions

Andy Hayes
Jun 24, 2016

The corresponding number in the complex plane is ( 4 + 7 3 ) + i ( 7 − 4 3 ) \left(4+7\sqrt{3}\right)+i\left(7-4\sqrt{3}\right) .

The corresponding number for the rotation is e π i / 3 = 1 2 + i 3 2 e^{\pi i/3}=\frac{1}{2}+i\frac{\sqrt{3}}{2} .

To achieve the rotation image in the complex plane, multiply these numbers.

( 3 2 ( 4 + 7 3 ) + i ( 7 − 4 3 ) ) ( 1 2 + i 3 2 ) = 8 + 14 i \left(\vphantom{\frac{\sqrt{3}}{2}}\left(4+7\sqrt{3}\right)+i\left(7-4\sqrt{3}\right)\right)\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)=8+14i

This corresponds to the point ( 8 , 14 ) (8,14) in the coordinate plane. Therefore, a + b = 22 a+b=\boxed{22} .

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Why does multiplication give us the rotation image?

Atomsky Jahid - 4 years, 11 months ago

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It's one of the interesting properties of complex numbers. For every complex number a + b i a+bi , there is a magnitude, r = a 2 + b 2 r=\sqrt{a^2+b^2} , and an angle, θ = arctan b a \theta=\text{arctan}\frac{b}{a} . In polar form , the complex number is written as r e i θ re^{i\theta} . You can put the complex number back into standard form using Euler's formula : r e i θ = r cos ⁡ θ + i r sin ⁡ θ re^{i\theta}=r\cos{\theta}+ir\sin{\theta}

Suppose there are two complex numbers r 1 e i θ 1 r_1e^{i\theta_1} and r 2 e i θ 2 r_2e^{i\theta_2} . Their product is:

r 1 e i θ 1 × r 2 e i θ 2 = r 1 r 2 e i ( θ 1 + θ 2 ) = r 1 r 2 cos ⁡ ( θ 1 + θ 2 ) + i r 1 r 2 sin ⁡ ( θ 1 + θ 2 ) \begin{aligned} r_1e^{i\theta_1}\times r_2e^{i\theta_2}&=r_1r_2e^{i(\theta_1+\theta_2)} \\ &=r_1r_2\cos(\theta_1+\theta_2)+ir_1r_2\sin(\theta_1+\theta_2) \end{aligned}

The consequence of multiplying complex numbers is that their angles are added together. Thus, a rotation is achieved.

Andy Hayes - 4 years, 11 months ago
Tom Engelsman
Feb 24, 2021

Using the rotation transformation matrix in R 2 \mathbb{R^2} :

[ cos ⁡ θ − sin ⁡ θ sin ⁡ θ cos ⁡ θ ] \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}

the require image point is computed per:

[ cos ⁡ ( π / 3 ) − sin ⁡ ( π / 3 ) sin ⁡ ( π / 3 ) cos ⁡ ( π / 3 ) ] ⋅ [ 4 + 7 3 7 − 4 3 ] = [ 8 14 ] = [ a b ] . \begin{bmatrix} \cos(\pi/3) & -\sin(\pi/3) \\ \sin(\pi/3) & \cos(\pi/3) \end{bmatrix} \cdot \begin{bmatrix} 4+7\sqrt{3} \\ 7 -4\sqrt{3} \end{bmatrix} = \begin{bmatrix} 8 \\ 14 \end{bmatrix} =\begin{bmatrix} a \\ b \end{bmatrix} .

Thus, a + b = 22 . a+b=\boxed{22}.

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