1.15 Escreva na forma a+bi os números abaixo.Use apenas a primeira raiz do número complexo (aquela em que k=0 onde for o caso. (1+sqrt (3)i)^1/2+(7+2i) b) (sqrt (3)-i)^1/2-(3-i)(2-3i)^-1

Vamos corrigir e detalhar os cálculos para obter as respostas corretas.<br /><br />### a) \((1+\sqrt{3}i)^{1/2} + (7+2i)\)<br /><br />Primeiro, vamos encontrar a raiz quadrada de \(1+\sqrt{3}i\).<br /><br />1. **Convertendo \(1+\sqrt{3}i\) para a forma polar:**<br /><br /> \[<br /> 1+\sqrt{3}i = r(\cos \theta + i \sin \theta)<br /> \]<br /><br /> Onde \(r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2\) e \(\theta = \arctan\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}\).<br /><br /> Assim, \(1+\sqrt{3}i = 2 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)\).<br /><br />2. **Encontrando a raiz quadrada:**<br /><br /> \[<br /> \sqrt{1+\sqrt{3}i} = \sqrt{2} \left( \cos \frac{\pi/6 + k\pi}{2} + i \sin \frac{\pi/6 + k\pi}{2} \right)<br /> \]<br /><br /> Para \(k = 0\):<br /><br /> \[<br /> \sqrt{1+\sqrt{3}i} = \sqrt{2} \left( \cos \frac{\pi/6}{2} + i \sin \frac{\pi/6}{2} \right) = \sqrt{2} \left( \cos \frac{\pi}{12} + i \sin \frac{\pi}{12} \right)<br /> \]<br /><br />3. **Calculando os valores:**<br /><br /> \[<br /> \cos \frac{\pi}{12} \approx 0.9659, \quad \sin \frac{\pi}{12} \approx 0.2588<br /> \]<br /><br /> Então,<br /><br /> \[<br /> \sqrt{1+\sqrt{3}i} \approx \sqrt{2} \left( 0.9659 + 0.2588i \right) \approx 1.414 \left( 0.9659 + 0.2588i \right) \approx 1.366 + 0.366i<br /> \]<br /><br />4. **Somando com \(7+2i\):**<br /><br /> \[<br /> (1.366 + 0.366i) + (7 + 2i) = 8.366 + 2.366i<br /> \]<br /><br />Portanto, a resposta correta é:<br /><br />\[<br />(a) \quad 8.366 + 2.366i<br />\]<br /><br />### b) \((\sqrt{3}-i)^{1/2} - (3-i)(2-3i)^{-1}\)<br /><br />1. **Convertendo \(\sqrt{3}-i\) para a forma polar:**<br /><br /> \[<br /> \sqrt{3}-i = 2 \left( \cos \left(-\frac{\pi}{6}\right) + i \sin \left(-\frac{\pi}{6}\right) \right)<br /> \]<br /><br />2. **Encontrando a raiz quadrada:**<br /><br /> \[<br /> \sqrt{\sqrt{3}-i} = \sqrt{2} \left( \cos \left(-\frac{\pi/6 + k\pi}{2}\right) + i \sin \left(-\frac{\pi/6 + k\pi}{2}\right) \right)<br /> \]<br /><br /> Para \(k = 0\):<br /><br /> \[<br /> \sqrt{\sqrt{3}-i} = \sqrt{2} \left( \cos \left(-\frac{\pi/12}\right) + i \sin \left(-\frac{\pi/12}\right) \right) = \sqrt{2} \left( \cos \left(-\frac{\pi}{12}\right) + i \sin \left(-\frac{\pi}{12}\right) \right)<br /> \]<br /><br />3. **Calculando os valores:**<br /><br /> \[<br /> \cos \left(-\frac{\pi}{12}\right) \approx 0.9659, \quad \sin \left(-\frac{\pi}{12}\right) \approx -0.2588<br /> \]<br /><br /> Então,<br /><br /> \[<br /> \sqrt{\sqrt{3}-i} \approx \sqrt{2} \left( 0.9659 - 0.2588i \right) \approx 1.414 \left( 0.9659