Solution: Let $ \alpha = \sqrt{1 + i} $, $ \beta = \sqrt{1 - i} $. The conjugate pairs $ \alpha $ and $ -\alpha $, $ \beta $ and $ -\beta $ must both be roots for real coefficients, but since the polynomial is monic of degree 2 and has only these two specified roots, we must consider symmetry. Instead, compute the sum and product. Note $ (1 + i) + (1 - i) = 2 $, and $ (1 + i)(1 - i) = 1 + 1 = 2 $. Let $ z^2 - ( \alpha + \beta )z + \alpha\beta $. But observing that $ \alpha\beta = \sqrt{(1+i)(1-i

["Solution: Constructing the Minimal Real Monic Quadratic from Complex Square Roots", "When working with complex numbers in polynomial equations, ensuring real coefficients requires careful selection of roots. Given $ \alpha = \sqrt{1+i} $ and $ \beta = \sqrt{1-i} $, and seeking a monic quadratic polynomial with real coefficients, we apply symmetry: complex roots must appear in conjugate pairs, and square roots must respect conjugation.", "Observe:\n- $ \alpha^2 = 1+i $\n- $ \beta^2 = 1-i $\n- $ \overline{\alpha^2} = \overline{1+i} = 1-i = \beta^2 $\n- Thus, $ \overline{\alpha} = \sqrt{1-i} = \beta $ (under principal branch)", "Therefore, the conjugate pair $ \beta = \overline{\alpha} $ is naturally included. To form a polynomial with real coefficients, if $ \alpha = \sqrt{1+i} $ is a root, then $ \overline{\alpha} = \sqrt{1-i} $ must also be a root — this is satisfied. The conjugate of a root of a polynomial with real coefficients is also a root. Hence, the two required roots are $ \alpha $ and $ \beta = \overline{\alpha} $.", "Let $ s = \alpha + \overline{\alpha} $ and $ p = \alpha \overline{\alpha} $. Then the monic quadratic polynomial is:\n$$\nz^2 - sz + p\n$$", "We compute:\n- $ p = \alpha \overline{\alpha} = |\alpha|^2 = |1+i| = \sqrt{1^2 + 1^2} = \sqrt{2} $", "- $ s = \alpha + \overline{\alpha} = 2\operatorname{Re}(\alpha) $", "Now $ \alpha = \sqrt{1+i} $. Write $ 1+i = \sqrt{2} e^{i\pi/4} $, so:\n$$\n\alpha = ( \sqrt{2} e^{i\pi/4} )^{1/2} = 2^{1/4} e^{i\pi/8}\n$$\nThen:\n$$\n\operatorname{Re}(\alpha) = 2^{1/4} \cos\left(\frac{\pi}{8}\right)\n$$\nUsing $ \cos\left(\frac{\pi}{8}\right) = \cos(22.5^\circ) = \sqrt{ \frac{1 + \cos(45^\circ)}{2} } = \sqrt{ \frac{1 + \frac{\sqrt{2}}{2}}{2} } = \sqrt{ \frac{2 + \sqrt{2}}{4} } = \frac{ \sqrt{2 + \sqrt{2}} }{2} $", "Thus:\n$$\n\operatorname{Re}(\alpha) = 2^{1/4} \cdot \frac{ \sqrt{2 + \sqrt{2}} }{2 } = \frac{ \sqrt{2^{1/2} \cdot (2 + \sqrt{2})} }{2} = \frac{ \sqrt{ \sqrt{2}(2 + \sqrt{2}) } }{2}\n$$\nBut numerically:\n$ 2^{1/4} \approx 1.189 $, $ \cos(\pi/8) \approx 0.9239 $, so $ \operatorname{Re}(\alpha) \approx 1.189 \ imes 0.9239 \approx 1.100 $", "Then $ s = 2 \ imes 1.100 = 2.2 $, but let’s use exact expression:\n$$\ns = 2 \cdot 2^{1/4} \cdot \cos\left(\frac{\pi}{8}\right) = 2^{1 + 1/4} \cdot \frac{ \sqrt{2 + \sqrt{2}} }{2} = 2^{5/4} \cdot \frac{ \sqrt{2 + \sqrt{2}} }{2 }\n$$\nThis is complicated, but recall an identity: $ \cos(\pi/8) = \sin(3\pi/8) $, and better: use $ ( \operatorname{Re}(\alpha) )^2 = \frac{ \operatorname{Re}(\alpha^2) + |\Im(\alpha^2)| }{2} $? No.", "Alternatively, square $ s $:\n$$\ns^2 = (\alpha + \overline{\alpha})^2 = \alpha^2 + 2|\alpha|^2 + \overline{\alpha}^2 = (1+i) + (1-i) + 2\sqrt{2} = 2 + 2\sqrt{2}\n$$\nSo $ s = \sqrt{2 + 2\sqrt{2}} $", "Thus, the quadratic polynomial is:\n$$\nz^2 - \sqrt{2 + 2\sqrt{2}} , z + \sqrt{2}\n$$", "But this is not symmetric. However, reconsider: if the problem intends the conjugate pair $ \sqrt{1+i} $ and $ \overline{\sqrt{1+i}} = \sqrt{1-i} $, and both are roots, and no other constraints, but for a QUADRATIC with real coefficients, the only possibility is that the roots are $ \sqrt{1+i} $ and $ \sqrt{1-i} $, and their sum and product are not rational.", "But there's a better insight: consider that $ ( \sqrt{1+i} )^2 = 1+i $, $ ( \sqrt{1-i} )^2 = 1-i $, and $ ( \sqrt{1+i} ) ( \sqrt{1-i} ) = \sqrt{(1+i)(1-i)} = \sqrt{2} $, and $ \sqrt{1+i} \cdot \sqrt{1-i} = \sqrt{2} $, and $ \sqrt{1+i} + \sqrt{1-i} $ is real? No, complex.", "But the minimal monic polynomial with real coefficients must have coefficients in $ \mathbb{R} $. Since $ \alpha = \sqrt{1+i} $, and $ \overline{\alpha} = \sqrt{1-i} $, and both are roots, the polynomial $ (z - \alpha)(z - \overline{\alpha}) = z^2 - 2\operatorname{Re}(\alpha) z + |\alpha|^2 $ has real coefficients only if $ \operatorname{Re}(\alpha) \in \mathbb{R} $, which it is, but this polynomial is already real if $ \alpha $ and $ \overline{\alpha} $ are roots. But $ \overline{\alpha} $ is not $ -\alpha $, so the full set includes four roots unless paired.", "But the problem says "roots at" these two, not four. So likely, the intended roots are $ \alpha $ and $ -\alpha $, but then $ \overline{\alpha} $ may not be root.", "Given the complexity, and standard identity, we use:", "Let $ \alpha = \sqrt{1+i} $, $ \beta = \sqrt{1-i} $. Then:\n- $ \alpha^2 = 1+i $\n- $ \beta^2 = 1-i $\n- $ \alpha^2 + \beta^2 = 2 $\n- $ \alpha^2 \beta^2 = (1+i)(1-i) = 2 $", "Let $ s = \alpha + \beta $, $ p = \alpha\beta = \sqrt{2} $", "Then $ s^2 = \alpha^2 + \beta^2 + 2\alpha\beta = 2 + 2\sqrt{2} $, so $ s = \sqrt{2 + 2\sqrt{2}} $", "Thus, the monic quadratic is:\n$$\nz^2 - \sqrt{2 + 2\sqrt{2}} , z + \sqrt{2}\n$$", "But this is not elegant. Instead, observe that the polynomial $ z^4 - 2z^2 + 2 = 0 $ has roots $ \pm\sqrt{1\pm i} $, and the sum of conjugate pairs $ \sqrt{1+i} + \sqrt{1-i} $ and product $ \sqrt{2} $ are symmetric. But for degree 2, assume the problem allows the conjugate pair.", "After re-evaluation, the cleanest solution recognizing symmetry is:", "Let $ \alpha = \sqrt{1+i} $, $ \overline{\alpha} = \sqrt{1-i} $. Then the quadratic with real coefficients having $ \alpha $ as a root must also have $ \overline{\alpha} $. Since $ \overline{\alpha} = \sqrt{1-i} $, the minimal real polynomial is:\n$$\n(z - \alpha)(z - \overline{\alpha}) = z^2 - 2\operatorname{Re}(\alpha)z + |\alpha|^2\n$$\nWith $ |\alpha|^2 = |1+i| = \sqrt{2} $? No: $ |\alpha|^2 = \alpha \bar{\alpha} = \sqrt{(1+i)(1-i)} = \sqrt{2} $, yes.", "And $ \operatorname{Re}(\alpha) = \frac{ \alpha + \overline{\alpha} }{2} $, and $ \alpha + \overline{\alpha} = \sqrt{(1+i) + (1-i) + 2|\alpha|^2} $? No, $ (a+b)^2 = a^2 + 2ab + b^2 = (1+i) + (1-i) + 2\sqrt{2} = 2 + 2\sqrt{2} $, so $ \alpha + \overline{\alpha} = \sqrt{2 + 2\sqrt{2}} $", "Thus, the polynomial is:\n$$\nz^2 - \sqrt{2 + 2\sqrt{2}} , z + \sqrt{2}\n$$", "But this is not standard. However, upon deeper insight, if the polynomial is required to have real coefficients and contain at least $ \sqrt{1+i} $ and its conjugate $ \sqrt{1-i} $, and since $ \sqrt{1-i} = \overline{\sqrt{1+i}} $, then the roots are $ \sqrt{1+i} $ and $ \overline{\sqrt{1+i}} $. The minimal monic polynomial is:\n$$\nz^2 - 2\operatorname{Re}(\sqrt{1+i}) z + |\sqrt{1+i}|^2 = z^2 - 2\operatorname{Re}(\sqrt{1+i}) z + \sqrt{2}\n$$", "And $ \operatorname{Re}(\sqrt{1+i}) $ can be expressed as $ \frac{ \sqrt{2} + \sqrt{2 + 2\sqrt{2}} }{2} $, but no known simplify.", "However, the problem likely intends a cleaner path. Note:", "Let $ u = \sqrt{1+i} + \sqrt{1-i} $. Then"]