Ejercicios de repaso

1. Hallar las soluciones reales de las ecuaciones
a) $(3x-1)(2+i)+(x-iy)(1+2i)=5+6i$

Desarrollo

$(3x-1)(2+i)+(x-iy)(1+2i)=5+6i$
$(3x-1)(2)+(3x-1)i+(x+2y)+(2x-y)i=5+6i$
$6x-2+3xi-i+x+2y+2xi-yi=5+6i$
$(7x-2+2y)+(5x-1-y)i=5+6i$

$\{_{5x-1-y=6}^{7x-2+y=5}  $ $\rightarrow$  $\{_{5x-y=7}^{7x+2y=7}$

$\boxed{x=\frac{21}{17}}$   $\boxed{y=\frac{-14}{17}}$

b) $(x-iy)(a+ib)=i^5$ donde $a,b \in \mathbb{R}$ y $\vert a \vert \not=\vert b \vert$

Desarrollo

$(x-iy)(a-ib)=i^5$
$(ax-by)+(-bx-ay)i=i$

$\{_{bx+ay=-1}^{ax-by=0}$

$x=\frac{b}{a}\cdot y$

$\rightarrow b(\frac{b}{a}\cdot y)+ay=-1$
$\frac{b^2}{a}\cdot y+ay=-1$
$y(\frac{b^2+a^2}{a})=-1$
$\boxed{y=-\frac{a}{a^2+b^2}}$

$x=\frac{b}{a}(-\frac{a}{a^2+b^2})$
$\boxed{x=-\frac{b}{a^2+b^2}}$

2. Resolver  $\{_{(2+i)x+(2-i)y=2i}^{(1+i)x-iy=2}$

Desarrollo

Usando la regla de Cramer tenemos:

$
\begin{pmatrix}
i+i  &  -i  \\
2+i  &  2-i  \\
\end{pmatrix}
$ $\cdot$  $
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
$ $=$  $
\begin{pmatrix}
2 \\
2i \\
\end{pmatrix}
$

$\vert A \vert=$ $\left| \begin{array}{crl}
1+i  & -i  \\
2+i  &  2-i
\end{array}\right| $ $=2+3i$

$\vert A_x \vert=$ $\left|  \begin{array}{crl}
2  &  -i  \\
2i  &  2-i
\end{array}\right| $ $=2-2i$

$\vert A_y \vert=$ $\left| \begin{array}{crl}
1+i  &  2  \\
2+i  &  2i
\end{array}\right| $ $=-6$

$x=\frac{\vert A_x\vert}{\vert A\vert}=\frac{2-2i}{2+3i}=\frac{-2-10i}{13}$
$\boxed{x=-\frac{2}{13}-\frac{10}{13}i}$


$y=\frac{\vert A_y\vert}{\vert A\vert}=\frac{-12+18i}{13}$
$\boxed{y=-\frac{12}{13}+\frac{18}{13}i}$

3. Comprobar que
a) $\frac{5}{(1-i)(2-i)(3-i)}=\frac{i}{2}$

Desarrollo

$\frac{5}{(i-i)(2-i)(3-i)}=\frac{5}{(1-3i)(3-i)}=\frac{5}{-10i}=\boxed{\frac{i}{2}}$

b) $\frac{1+2i}{3-4i}+\frac{2-i}{5i}=-\frac{2}{5}$

Desarrollo

$\frac{1+2i}{3-4i}+\frac{2-i}{5i}=\frac{(1+2i)(3-4i)}{25}-\frac{(2-i)i}{5}=\frac{-5+10i}{25}-\frac{1+2i}{5}=\frac{-1+2i}{5}-\frac{1+2i}{5}=\boxed{-\frac{2}{5}}$

4. Calcular $\frac{(1+i)^n}{(1-i)^{n-2}}$ para $n\in \mathbb{Z^+}$

Desarrollo

$\frac{(1+i)^n}{(1-i)^{n-2}}=\left( \frac{1+i}{1-i} \right)^n \cdot (1-i)^2=\left( \frac{(1+i)^2}{2} \right) ^n \cdot (1-i)^2=i^n \cdot (1-i)^2=1^n \cdot (-2i)=\boxed{-2i^{n+1}}$

5. Simplificar $(1+cos\theta+i sen\theta)^n+(1+cos\theta-i sen\theta)^n$

Desarrollo

$(1+cos\theta +i sen\theta)^n+(1+cos\theta -i sen\theta)^n$

Utilizando las identidades:

$2cos^2(\frac{\theta}{2})=1+cos\theta$
$sen\theta=2cos(\frac{\theta}{2}) sen(\frac{\theta}{2})$

La primera expresión se transforma en:
$(1+cos\theta+i sen\theta)^n+(1+cos\theta-i sen\theta)^n=[2cos^2 (\frac{\theta}{2})+i\cdot sen(\frac{\theta}{2})cos(\frac{\theta}{2})]^n+[2cos^2(\frac{\theta}{2})-i\cdot sen(\frac{\theta}{2})cos(\frac{\theta}{2})]^n$
$=2^n cos^n(\frac{\theta}{2})[(cos(\frac{\theta}{2})+i sen(\frac{\theta }{2}))^n+(cos(\frac{\theta}{2})-i sen(\frac{\theta}{2}))^n]$
$=2^ncos^n (\frac{\theta}{2})[cos(\frac{n\theta}{2})+i sen(\frac{n\theta}{2})+cos(\frac{n\theta}{2})-i sen(\frac{n\theta}{2})]$
$=2^n cos^n (\frac{\theta}{2})[cos(\frac{n\theta}{2})]$
$=\boxed{2^{n+1}cos^n(\frac{\theta}{2}) cos(\frac{n\theta}{2})}$