Question

Determine o número complexo z tal que 2z̅− 1 = z̅− i.

b) Escreva z na forma trigonométrica.

Answer 1

Explicação passo-a-passo:

Seja $\sf z=a+bi$. Assim, $\sf \overline{z}=a-bi$

$\sf 2\overline{z}-1=\overline{z}-i$

$\sf 2\cdot(a-bi)-1=a-bi-i$

$\sf 2a-2bi-1=a-(b+1)i$

$\sf 2a-1-2bi=a-(b+1)i$

Assim:

• $\sf 2a-1=a$

$\sf 2a-a=1$

$\sf \red{a=1}$

• $\sf 2b=b+1$

$\sf 2b-b=1$

$\sf \red{b=1}$

Então, $\sf \red{z=1+i}$

$\sf \rho=\sqrt{1^2+1^2}$

$\sf \rho=\sqrt{1+1}$

$\sf \rho=\sqrt{2}$

Temos que:

• $\sf sen~\theta=\dfrac{1}{\sqrt{2}}~\Rightarrow~sen~\theta=\dfrac{\sqrt{2}}{2}$

• $\sf cos~\theta=\dfrac{1}{\sqrt{2}}~\Rightarrow~cos~\theta=\dfrac{\sqrt{2}}{2}$

Logo, $\sf \theta=\dfrac{\pi}{4}~rad$

A forma trigonométrica é:

$\sf z=|~z~|\cdot(cos~\theta+i\cdot sen~\theta)$

$\sf \red{z=\sqrt{2}\cdot\Big[cos~\Big(\dfrac{\pi}{4}\Big)+i\cdot sen~\Big(\dfrac{\pi}{4}\Big)\Big]}$