Question

Passe para forma trigonométrica os complexos:
a) Z = 1
b) Z= -1
c) Z= -1 + i

Answer 1

Explicação passo-a-passo:

a)

$\sf \rho=\sqrt{1^2+0^2}$

$\sf \rho=\sqrt{1+0}$

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

$\sf \rho=1$

Temos que:

$\sf sen~\theta=\dfrac{0}{1}~\Rightarrow~sen~\theta=0$

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

Assim, $\sf \theta=0$

Logo:

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

$\sf z=1\cdot(cos~0+i\cdot sen~0)$

b)

$\sf \rho=\sqrt{(-1)^2+0^2}$

$\sf \rho=\sqrt{1+0}$

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

$\sf \rho=1$

Temos que:

$\sf sen~\theta=\dfrac{0}{1}~\Rightarrow~sen~\theta=0$

$\sf cos~\theta=\dfrac{-1}{1}~\Rightarrow~cos~\theta=-1$

Assim, $\sf \theta=\pi~rad$

Logo:

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

$\sf z=1\cdot(cos~\pi+i\cdot sen~\pi)$

c)

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

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

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

Temos que:

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

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

Assim, $\sf \theta=\dfrac{3\pi}{4}~rad$

Logo:

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

$\sf z=\sqrt{2}\cdot\Big[cos~\Big(\dfrac{3\pi}{4}\Big)+i\cdot sen~\Big(\dfrac{3\pi}{4}\Big)\Big]$