Usuario discusión:Cecilia Carrizosa Muñoz

1.5 Sean w,z ∈ C. Demuestre los siguientes incisos:

Sean $z=a+ib\qquad y\qquad w=c+id$ entonces:

(1) ${\overline {\overline {z}}}=z$

Solución:

${\overline {\overline {z}}}={\overline {\overline {(a+ib)}}}={\overline {(a-ib)}}=a+ib=z$

$\therefore {\overline {\overline {z}}}=z$

(2)${\overline {z+w}}={\overline {z}}+{\overline {w}}$

Solución:

${\overline {z+w}}={\overline {(a+ib)+(c+id)}}={\overline {(a+c)+(b+d)i}}=(a+c)-(b+d)i=$

$=(a-bi)+(c-di)={\overline {z}}+{\overline {w}}$

$\therefore {\overline {z+w}}={\overline {z}}+{\overline {w}}$

(3)${\overline {zw}}={\overline {z}}*{\overline {w}}$

Solución:

${\overline {zw}}={\overline {(a+ib)(c+id)}}={\overline {(ac-bd)+(ad+bc)i}}=(ac-bd)-(ad+bc)i=ac-iad+bd-icd=$

$=ac-iad+i^{2}bd-icb=a(c-id)-ib(c-id)=(a-ib)(c-id)={\overline {z}}*{\overline {w}}$

$\therefore {\overline {zw}}={\overline {z}}*{\overline {w}}$

(6)$z\in {R}\Longleftrightarrow {\overline {z}}=z$ (es decir, un número complejo es un número real si y sólo si es igual a su conjugado).

Solución:

Sea $z=a+0i=a$, entonces:

${\overline {z}}={\overline {a+0i}}=a-0i=a$

$\therefore {\overline {z}}=z$

1.9 Haga las operaciones indicadas y al final exprese el resultado en la forma a+bi

(a)$\qquad (3+2i)(5-3i)$

$\qquad (3+2i)(5-3i)=(3*5)+(2*-3)i=15-6i$

(b)$\qquad (5+7i)-(4-2i)$

$\qquad (5+7i)-(4-2i)=-20+14i$

(c)$\qquad 3i-(-7+2i)$

$\qquad 3i-(-7+2i)=7+(3-2)i=7+i$

(d)$5+{\bigg (}{\frac {1}{2}}-3i{\bigg )}$

$5+{\bigg (}{\frac {1}{2}}-3i{\bigg )}={\frac {11}{2}}-3i$

(e)$\qquad (2+3i)(4-2i)$

$\qquad (2+3i)(4-2i)=14+8i$

(f)$\qquad {\frac {3+4i}{5+2i}}$

${\frac {3+4i}{5+2i}}={\bigg (}{\frac {3+4i}{5+2i}}{\bigg )}{\bigg (}{\frac {5-2i}{5-2i}}{\bigg )}={\frac {23+14i}{29}}={\frac {23}{29}}+{\frac {14}{29}}i$

(g)${\frac {4}{2-3i}}$

${\frac {4}{2-3i}}={\frac {4}{2-3i}}{\bigg (}{\frac {2+3i}{2+3i}}{\bigg )}={\frac {8+12i}{13}}={\frac {8}{13}}+{\frac {12}{13}}i$

(h)${\frac {(2+i)(3+2i)}{1-i}}$

${\frac {(2+i)(3+2i)}{1-i}}={\frac {4+7i}{1-i}}={\frac {4+7i}{1-i}}{\bigg (}{\frac {1+i}{1+i}}{\bigg )}={\frac {11+3i}{2}}={\frac {11}{2}}+{\frac {3}{2}}i$