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China Mathematical Competition

China algebra

Problem

Let complex number

z = (a + cos θ) + (2 a - sin θ) i

. If

∣ z ∣ \leq 2

for any

θ \in R

, then the range of real number

a

is ______.

Solution

By the definition given above, we have, for any

θ \in R

,

∣ z ∣ \leq 2 \Leftrightarrow (a + cos θ)^{2} + (2 a - sin θ)^{2} \leq 4 \Leftrightarrow 2 a (cos θ - 2 sin θ) \leq 3 - 5 a^{2} \Leftrightarrow - 25 a sin (θ - φ) \leq 3 - 5 a^{2} \Rightarrow 25 ∣ a ∣ \leq 3 - 5 a^{2} \Rightarrow ∣ a ∣ \leq \frac{5}{5} .

So the range of

a

is

[- \frac{5}{5}, \frac{5}{5}]

.

Final answer

[-√5/5, √5/5]

Techniques

Complex numbersLinear and quadratic inequalities