Absolute value and complex number

Algebra Level 4

1512 + k i 2688 + k i = 3 4 \large \dfrac{|1512+ki|}{|2688+ki|} = \dfrac{3}{4}

Find k |k| that satisfies the equation above.

Notations :


The answer is 2016.

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1 solution

Tommy Li
Jul 14, 2016

1512 + k i 2688 + k i = 3 4 \dfrac{|1512+ki|}{|2688+ki|} = \dfrac{3}{4}

151 2 2 + k 2 268 8 2 + k 2 = 3 4 \dfrac{\sqrt{1512^2+k^2}}{\sqrt{2688^2+k^2}} = \dfrac{3}{4}

151 2 2 + k 2 268 8 2 + k 2 = 9 16 \dfrac{1512^2+k^2}{2688^2+k^2} = \dfrac{9}{16}

16 ( 151 2 2 + k 2 ) = 9 ( 268 8 2 + k 2 ) 16(1512^2+k^2)=9(2688^2+k^2)

7 k 2 = 28449792 7k^2=28449792

k 2 = 4064256 k^2=4064256

k = 2016 k=2016 or k = 2016 k=-2016

k |k| = 2016

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