[Toán 12] Tìm max|z|

N

nguyenbahiep1

$dk: z \not =0, z= a+bi \\ \\ |\frac{z^2+1}{z}| = 1 \\ \\ |z^2+1| = |z| \Rightarrow (a^2-b^2+1)^2 + 4a^2b^2 = a^2+b^2 \\ \\ \Rightarrow a^4+b^4+2a^2b^2+a^2-3b^2+1 = 0 \\ \\ |z| = a^2+b^2 = k \\ \\ b^2 = k-a^2 \Rightarrow 4a^2+k^2-3k+1 = 0 \Rightarrow k^2-3k+1 \leq 0 \Rightarrow k \leq \frac{3+\sqrt{5}}{2}\\ \\ Max_{|z|}= \frac{3+\sqrt{5}}{2}$
 
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