Question

determine a raiz quadrada dos complexos a)-5+12i b) 4i c) 4+3i d) 1-iraizde 3

Answer 1

Dado um número complexo

$z=a\pm bi$

onde $a,\,b\in \mathbb{R}\;\;\text{ e }\;\;b\geq 0$


a raiz quadrada de $z$ pode ser obtida pela fórmula:

$\sqrt{a\pm bi}=\sqrt{\frac{\sqrt{a^{2}+b^{2}}+a}{2}}\pm i\sqrt{\frac{\sqrt{a^{2}+b^{2}}-a}{2}}$


a) $-5+12i\;\;\;\Rightarrow\;\;\left\{ \begin{array}{l}a=-5\\b=12 \end{array} \right.$


Calculando o módulo:

$\sqrt{a^{2}+b^{2}}=\sqrt{5^{2}+12^{2}}\\ \\ \sqrt{a^{2}+b^{2}}=\sqrt{25+144}\\ \\ \sqrt{a^{2}+b^{2}}=\sqrt{169}\\ \\ \sqrt{a^{2}+b^{2}}=13$


Substituindo na fórmula da raiz quadrada, temos:

$\sqrt{-5+12i}=\sqrt{\frac{13+(-5)}{2}}+i\sqrt{\frac{13-(-5)}{2}}\\ \\ \sqrt{-5+12i}=\sqrt{\frac{13-5}{2}}+i\sqrt{\frac{13+5}{2}}\\ \\ \sqrt{-5+12i}=\sqrt{\frac{8}{2}}+i\sqrt{\frac{18}{2}}\\ \\ \sqrt{-5+12i}=\sqrt{4}+i\sqrt{9}\\ \\ \boxed{\begin{array}{c}\sqrt{-5+12i}=2+3i \end{array}}$


b) $4i\;\;\;\Rightarrow\;\;\left\{ \begin{array}{l}a=0\\b=4 \end{array} \right.$


Calculando o módulo:

$\sqrt{a^{2}+b^{2}}=\sqrt{0^{2}+4^{2}}\\ \\ \sqrt{a^{2}+b^{2}}=\sqrt{16}\\ \\ \sqrt{a^{2}+b^{2}}=4$


Substituindo na fórmula da raiz quadrada, temos:

$\sqrt{4i}=\sqrt{\frac{4+0}{2}}+i\sqrt{\frac{4-0}{2}}\\ \\ \sqrt{4i}=\sqrt{\frac{4}{2}}+i\sqrt{\frac{4}{2}} \\ \\ \boxed{\begin{array}{c}\sqrt{4i}=\sqrt{2}+\sqrt{2}\,i \end{array}}$


c) $4+3i\;\;\;\Rightarrow\;\;\left\{ \begin{array}{l}a=4\\b=3 \end{array} \right.$


Calculando o módulo:

$\sqrt{a^{2}+b^{2}}=\sqrt{4^{2}+3^{2}}\\ \\ \sqrt{a^{2}+b^{2}}=\sqrt{16+9}\\ \\ \sqrt{a^{2}+b^{2}}=\sqrt{25}\\ \\ \sqrt{a^{2}+b^{2}}=5$


Substituindo na fórmula da raiz quadrada, temos:

$\sqrt{4+3i}=\sqrt{\frac{5+4}{2}}+i\sqrt{\frac{5-4}{2}}\\ \\ \sqrt{4+3i}=\sqrt{\frac{9}{2}}+i\sqrt{\frac{1}{2}}\\ \\ \sqrt{4+3i}=\sqrt{\frac{9\cdot 2}{2\cdot 2}}+i\sqrt{\frac{1\cdot 2}{2\cdot 2}}\\ \\ \sqrt{4+3i}=\sqrt{\frac{9\cdot 2}{4}}+i\sqrt{\frac{2}{4}}\\ \\ \sqrt{4+3i}=\frac{\sqrt{9}\cdot\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} \\ \\ \boxed{\begin{array}{c}\sqrt{4+3i}=\frac{3\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\,i \end{array}}$


d) $1-\sqrt{3}\,i\;\;\;\Rightarrow\;\;\left\{ \begin{array}{l}a=1\\b=\sqrt{3} \end{array} \right.$


Calculando o módulo:

$\sqrt{a^{2}+b^{2}}=\sqrt{1+(\sqrt{3})^{2}}\\ \\ \sqrt{a^{2}+b^{2}}=\sqrt{1+3}\\ \\ \sqrt{a^{2}+b^{2}}=\sqrt{4}\\ \\ \sqrt{a^{2}+b^{2}}=2$


Substituindo na fórmula da raiz quadrada, temos:

$\sqrt{1-\sqrt{3}\,i}=\sqrt{\frac{2+1}{2}}-i\sqrt{\frac{2-1}{2}}\\ \\ \sqrt{1-\sqrt{3}\,i}=\sqrt{\frac{3}{2}}-i\sqrt{\frac{1}{2}}\\ \\ \sqrt{1-\sqrt{3}\,i}=\sqrt{\frac{3\cdot 2}{2\cdot 2}}-i\sqrt{\frac{1\cdot 2}{2\cdot 2}}\\ \\ \sqrt{1-\sqrt{3}\,i}=\sqrt{\frac{6}{4}}-i\sqrt{\frac{2}{4}}\\ \\ \boxed{\begin{array}{c}\sqrt{1-\sqrt{3}\,i}=\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2}\,i \end{array}}$