Question

Exercitiile 6 sau 7 va rog frumos care știți......

Answer 1

Răspuns:

Explicație pas cu pas:

Ex6

a) $z=\frac{a+2i}{2+ai}=\frac{(a+2i)(2-ai)}{(2+ai)(2-ai)}=\frac{2a-a^{2}i+4i+2a }{4+a^{2} }=\frac{4a+(4-a^{2})i }{4+a^{2} }=\frac{4a}{4+a^{2}}+\frac{4-a^{2} }{4+a^{2} }i  \\. Pentru ca z sa apartina multimii R tr. ca partea imaginara sa fie 0, dar asta e posibil pentru a=-2 sau a=2.\\    b) z=\frac{1}{a(1+i)+1-2i}=\frac{1}{(a+1)+(a-2)i}=\frac{(a+1)-(a-2)i}{((a+1)+(a-2)i)((a+1)-(a-2)i)}=\frac{(a+1)-(a-2)i}{(a+1)^{2}+(a-2)^{2}}=\frac{(a+1)-(a-2)i}{2a^{2}-2a+5 }=$

=$=\frac{a+1}{2a^{2}-2a+5}+\frac{2-a}{2a^{2}-2a+5}i\\ Egalam partea reala cu 2/5 si obtinem\\\frac{a+1}{2a^{2}-2a+5}=\frac{2}{5}\\ 4a^{2}-4a+10=5a+5,  4a^{2}-9a+5=0, de unde a=1 sau a=\frac{5}{4}$

Answer 2

$a)\quad z = \dfrac{a+2i}{2+ai} \\ \\ \text{Teorema: }\\ \text{Fie un numar z}\in \mathbb{C},\quad \text{daca }z = \overline{z} \Rightarrow z \in \mathbb{R}\\ \\z = \dfrac{a+2i}{2+ai}=\overline{z} = \dfrac{a-2i}{2-ai}\\ \\\Rightarrow\dfrac{a+2i}{2+ai}=\dfrac{a-2i}{2-ai} \Rightarrow(a+2i)(2-ai)= (2+ai)(a-2i) \Rightarrow\\ \\ \Rightarrow 2a-a^2i+4i+2a = 2a-4i+a^2i+2a \Rightarrow\\ \\ \Rightarrow -2a^2i+8i=0 \Rightarrow (-2a^2+8)i = 0\Rightarrow \boxed{a = \pm 2}$

$b)\quad z = \dfrac{1}{a(1+i)+1-2i} = \dfrac{1}{(a+1)+(a-2)i} = \\ \\ = \dfrac{(a+1)-(a-2)i}{(a+1)^2-[(a-2)i]^2} = \dfrac{a+1-ai+2i}{(a+1)^2+(a-2)^2} \\ \\ \Rightarrow \dfrac{a+1}{(a+1)^2+(a-2)^2} = \dfrac{2}{5} \Rightarrow 5(a+1) = 2(a+1)^2+2(a-2)^2 \\ \\ \Rightarrow 5a+5 = 2a^2+4a+2+2a^2-8a+8 \\ \\ \Rightarrow 4a^2-9a+5 = 0\\ \\ \Delta = 81 - 80 =1 \Rightarrow a\in \Big\{1,\dfrac{5}{4}\Big\}$