Carrotsticks
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Difficult Complex Numbers Question (1 Viewer)
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seanieg89
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Are you sure that is true? Consider the polynomial P(z)=z/2... = \sum_{k=0}^{n} a_k z^k $ be a polynomial with real coefficients $ a_k . \\\\ $Show that if $ |P(i)| < 1 $ , then there exists $ a,b \in \mathbb{R} $ such that $ P(a+ib)=0 $ and $ (a^2+b^2+1)^2 < 4b^2 + 1)
Carrotsticks
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Mistyped the question. The polynomial is meant to be monic and non-constant/non-zero, thanks.
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Nooblet94
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Just out of curiousity, what's Vieta's formulae?|<1$, we have that at least one of these factors has modulus less than one at $z=i$. From the triangle inequality, this is impossible for monic linear factors so we may assume $Q(z)=z^2+a_1z+a_0$ is such a factor.\\ Now $|Q(i)|^2=(a_0-1)^2+a_1^2<1\quad(*)\\$By the conjugate root theorem and Vieta's formulae, Q has roots $a\pm ib$ with $2a=-a_1,a^2+b^2=a_0.$ Applying the inequality (*) we get:\\$ (a_0+1)^2<1+4a_0-a_1^2\Rightarrow (a^2+b^2+1)^2<1+4b^2.\\ $Whence $a+ib$ is a root of $Q$ (and hence $P$) with the claimed properties.$)
RealiseNothing
what is that?It is Cowpea
Sum of roots, and products of roots for quadratics.
Carrotsticks
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Very nice solution! Here is my one, which is ~similar~ to yours:
 $ can be expressed as: $ \\\\ P(z) = \prod_{k=1}^{n} (z-z_k)$ where $ P(z_j) = 0, j \in \mathbb{N}, j \in [1,n] \\\\ $We are given that $ |P(i)| < 1 \\\\ \Rightarrow \left | \prod_{k=1}^{n} (z_k-i) \right | = \prod_{k=1}^{n} |z_k-i| < 1 \\\\ $Where $ z_k $ is a real root, the factor $ |z_k - i| = \sqrt{z_k ^2 + 1} \geq 1 $ so if we omit them from the product, the inequality is preserved:$ \\\\ \Rightarrow \prod_{k=1}^{n} (z_{c_k} - i) < 1 \\\\ $ where $ z_{c_k} $ is a complex root ie: $ \Im(z_{c_k}) \neq 0 \\\\ $However since the polynomial has real coefficients, the complex roots exist in conjugate pairs so we can express the product as the product of pairs: $ \\\\ \prod_k |z_{c_k} - i||z_{c_k}^* - i| < 1 \\\\ $So we are essentially guaranteed that the product of at LEAST one of these conjugate pairs is less than one. We will call this conjugate pair $ a \pm ib. \\\\ |a+ib + i| | a+ib - i| < 1 \\ )
^2 } \times \sqrt{a^2 + (b+1)^2} < 1 \\\\ \Rightarrow (a^2+b^2+1)^2 < 4b^2 + 1)