Question

hallar la raiz cuadrada de los complejos:
a)5+12i
b)1/(3+4i)

Answer 1

Tienes lo siguiente:
Buscas un número de la forma a + bi con a,b∈R que elevado al cuadrado sea el resultado que quieras:
$a) \\ \sqrt{5+12i}=a+bi\\ (\sqrt{5+12i})^2=(a+bi)^2\\5+12i=a^2+2abi-b^2 \\ \Rightarrow 5=a^2-b^2 \\ \Rightarrow 12i=2abi\Rightarrow ab=6\to a= \frac{6}{b} \\ 5=( \frac{6}{b} )^2-b^2= \frac{36}{b^2}-b^2 \\ \\ 5= \frac{36}{b^2}-b^2 \\ 5b^2=36-b^4 \\ b^4+5b^2-36=0 \\ \\ b^2= \frac{-5\pm \sqrt{5^2-4(1)(-36)} }{2(1)} = \frac{-5\pm \sqrt{169} }{2}= \frac{-5\pm13}{2} \\ b_1^2=(-5+13)/2=4\Rightarrow b= \sqrt{4}=2 \\ b_2^2=(-5-13)/2=-9\Rightarrow b= \sqrt{-9}=3i \\ a= \frac{6}{b}= \frac{6}{2}=3$

$\text{Como a y b son numeros reales, solo sustituyes b1} \\ \\ \boxed{a+bi=3+2i}$

$b) \\ (\frac{1}{3+4i})=(a+bi)^2=a^2+2abi-b^2 \\ \frac{1}{3+4i}\cdot \frac{3-4i}{3-4i}= \frac{3-4i}{9+16}= \frac{3}{25} - \frac{4i}{25} =a^2+2abi-b^2 \\ \\ \Rightarrow \frac{3}{25}=a^2-b^2 \\ \\ \Rightarrow -\frac{4i}{25}=2abi\Rightarrow -\frac{4}{50b}=a =- \frac{2}{25b} \\ \\ a^2-b^2= (- \frac{2}{25b} )^2-b^2= \frac{3}{25} \\ \\ \frac{4}{25^2b^2}-b^2= \frac{3}{25} \\ \\ 25^2b^2( \frac{4}{25^2b^2}-b^2= \frac{3}{25} )\Rightarrow 4-25^2b^4=75b^2 \\ 625b^4+75b^2-4=0$

$b^2= \frac{-75\pm \sqrt{75^2-4(625)(-4)} }{2(625)}= \frac{-75\pm \sqrt{15625} }{1250}= \frac{-75\pm125}{1250} \\ b_1^2=(-75+125)/1250=50/1250=1/25\to b= \sqrt{1/25}=1/5 \\ b_2^2=(-75-125)/1250=-200/1250=-4/25\to b= \sqrt{-4/25}=2i/5 \\ \\ \text{Tomas la solucion real b1=1/5} \\ a=- \frac{2}{25b}= - \frac{2}{25(1/5)}=- \frac{2}{5} \\ \\ \boxed{a+bi=- \frac{2}{5}+ \frac{i}{5} }$

Saludos!