Of je zo mag differentiëren is nog maar de vraag; daarvoor moet je eerst meer theorie opbouwen!

$z=r(cos\left(\phi \right)+isin\left(\phi \right))=r\cdot {\text{e}}^{\text{i}\phi }$

$z=\sqrt{8}\cdot {\text{e}}^{\mathrm{-0,25}\pi \text{i}}$

$z_1=\sqrt{\left(2\right)}\cdot {\text{e}}^{\mathrm{-0,25}\pi \text{i}}$

$z_2=\sqrt{\left(2\right)}\cdot {\text{e}}^{\mathrm{0,75}\pi \text{i}}$

$z_1\cdot z_2=2\cdot {\text{e}}^{\mathrm{0,5}\pi \text{i}}$

$z_1\cdot z_2=2\text{i}$

Duidelijk, toch?

$z=3{\text{e}}^{\text{i}}$

$z\approx \mathrm{-1,980}+0,282\text{i}$

Doen.

Doen.

$z\approx 5{\text{e}}^{\mathrm{-0,93}\text{i}}$

Doen.

$\left|z\right|=\sqrt{20}$ en $\text{arg}\left(z\right)\approx \mathrm{2,68}$

$z=\sqrt{20}\cdot {\text{e}}^{\mathrm{2,86}\text{i}}$

Doen, oefen met een medeleerling.

Doen.

Doen.

`z_1 * z_2 = r_1 * text(e)^(text(i)φ_1) * r_2 * text(e)^(text(i)φ_2) = r_1 * r_2 * text(e)^(text(i)(φ_1 + φ_2))`
`(z_1)/(z_2) = (r_1 * text(e)^(text(i)φ_1))/(r_2 * text(e)^(text(i)φ_2)) = (r_1)/(r_2) * text(e)^(text(i)(φ_1 - φ_2))`

Doen.

$-4-4\text{i}$

Klopt natuurlijk.

`(cos(φ) + text(i)sin(φ))^n = (text(e)^(text(i)φ))^n = text(e)^(n*text(i)φ)) = text(e)^(text(i)*nφ)) = cos(nφ) + text(i)sin(nφ)`

Doen.

${\left({\text{e}}^{\text{i}\phi }\right)}^n={\text{e}}^{\text{i}n\phi }$

$z=2{\text{e}}^{0\text{i}}$

$z=1{\text{e}}^{\mathrm{0,5}\pi \text{i}}$

$z=3{\text{e}}^{\mathrm{0,5}\pi \text{i}}$

$z=\sqrt{\left(2\right)}\cdot {\text{e}}^{\mathrm{-0,25}\pi \text{i}}$

$z=\sqrt{\left(2\right)}\cdot {\text{e}}^{\mathrm{0,75}\pi \text{i}}$

$z=\sqrt{\left(8\right)}\cdot {\text{e}}^{\mathrm{1,25}\pi \text{i}}$

$-128+128\text{i}$

$122+597\text{i}$ (niet afronden tussentijds!)

Doen.

${\text{e}}^{\text{i}\phi }+{\text{e}}^{\mathrm{-}\text{i}\phi }=2cos\left(\phi \right)$

$sin\left(\phi \right)=\mathrm{-0,5}i({\text{e}}^{\text{i}\phi }+{\text{e}}^{\mathrm{-}\text{i}\phi })$

$z=5+3\text{i}$

$z=\frac{3}{34}+\frac{5}{34}\text{i}$

$z=738-684\text{i}$

$z=\mathrm{-0,5}+\mathrm{0,5}\text{i}\sqrt{3}$