IME / ITA ⇒ (IME - 1977) Complexos

[Caduzin3445]

Seja z = a + bi (a, b 6 91).
Determine a e b tais que z2 = 3 - 4i.

Gab: 2 -i ou - 2 + i

[petras]

Caduzin3445,

[tex3](a+bi)^2 =a^2-b^2+2abi \\
Comparando: a^2-b^2 = 3(I) \\
2abi = -4i \implies ab = -2 \implies b = -\frac{2}{a}\\
Em(I):a ^2-\frac{4}{a^2}=3 \implies a^4-4-3a^2 = 0\\
\therefore a = -2 \implies b = 1 \vee a =2\ \implies b =- 1\\
\therefore z = 2-i \vee z = -2 +i\\
[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3](a+bi)^2 =a^2-b^2+2abi \\
Comparando: a^2-b^2 = 3(I) \\
2abi = -4i \implies ab = -2 \implies b = -\frac{2}{a}\\
Em(I):a ^2-\frac{4}{a^2}=3 \implies a^4-4-3a^2 = 0\\
\therefore a = -2 \implies b = 1 \vee a =2\ \implies b =- 1\\
\therefore z = 2-i \vee z = -2 +i\\
[/tex3]