Question

forma trigonométrica do número complexo z = 1 + i

Answer 1

Um número complexo na forma algébrica $z=a+bi$ pode ser representado na forma trigonométrica $z=p(\cos\theta+i\,\text{sen}\,\theta)$, em que:

$p=\sqrt{a^2+b^2};\qquad\qquad\cos\theta=\dfrac{a}{p};\qquad\qquad\text{sen}\,\theta=\dfrac{b}{p}.$


No complexo $z=1+i$, temos que $a=1$ e $b=1$.
Logo:

$p=\sqrt{a^2+b^2}\\\\p=\sqrt{(1)^2+(1)^2}\longrightarrow{p}=\sqrt2$

Por conseguinte:

$\begin{matrix}\cos\theta=\dfrac{a}{p}\\\\\cos\theta=\dfrac{1}{\sqrt2}\\\\\cos\theta=\dfrac{\sqrt2}{2}\\\\\therefore\theta=\dfrac{\pi}{4}+k2\pi,\quad\text{para}\ k\in\mathbb{N}\end{matrix}\qquad\qquad\begin{matrix}\text{sen}\,\theta=\dfrac{b}{p}\\\\\text{sen}\,\theta=\dfrac{1}{\sqrt2}\\\\\text{sen}\,\theta=\dfrac{\sqrt2}{2}\\\\\therefore\theta=\dfrac{\pi}{4}+k2\pi,\quad\text{para}\ k\in\mathbb{N}\end{matrix}$

Por fim, substituindo os valores na forma trigonométrica do número complexo:

$z=p(\cos\theta+i\,\text{sen}\,\theta)\\\\\\\boxed{\boxed{z=\sqrt2\left(\cos\!\left(\dfrac{\pi}{4}+k2\pi\right)+i\,\text{sen}\!\left(\dfrac{\pi}{4}+k2\pi\right)\right),\quad\text{para}\ k\in\mathbb{N}}}$

ou, se preferir:

$\boxed{\boxed{z=\sqrt2\cos\!\left(\dfrac{\pi}{4}+k2\pi\right)+i\sqrt2\,\text{sen}\!\left(\dfrac{\pi}{4}+k2\pi\right),\quad\text{para}\ k\in\mathbb{N}}}$



Note que:

$1+i=\sqrt2\cos\!\left(\dfrac{\pi}{4}+k2\pi\right)+i\sqrt2\,\text{sen}\!\left(\dfrac{\pi}{4}+k2\pi\right)$