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

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

`|z|=sqrt(2^2+2^2) = sqrt(8)` en `arg(z) = arctan(2/2) = 1/4pi` .

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

`|z|=sqrt(1^2+(text(-)1)^2) = sqrt(2)` en `arg(z) = arctan(text(-)1/1) = text(-)1/4 pi` .

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

`|z|=sqrt((text(-)1)^2+1^2) = sqrt(2)` en `arg(z) = arctan(text(-)1/1) = 3/4 pi` .

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

`z_1*z_2 = sqrt(2)*text(e)^(text(-)0,25pi text(i)) * sqrt(2)*text(e)^(0,75pi text(i)) = sqrt(2)*sqrt(2)*text(e)^(text(-)0,25pi text(i) + 0,75pi text(i)) = 2*text(e)^(0,5pi text(i))` .

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

`text(e)^(0,5pi text(i)) = text(i)` , want `text(i)` heeft lengte `1` en argument `0,5pi` .

Doen, stel de juiste waarden voor `a` en `b` in.

`bar(z)=3-4text(i)` , dus `|bar(z)| = 5` en `arg(bar(z))=arctan(text(-)4/3)~~text(-)0,93pi` .

Dus `bar(z)~~5text(e)^(text(-)0,93text(i))` .

Doen, `|z| ~~ 4,47` .

`|z|=sqrt((text(-)4)^2+2^2)=sqrt(20)` en `arg(z)=pi+arctan(2/4)~~2,68` .

$z=\sqrt{20}\cdot {\text{e}}^{\mathrm{2,68}\text{i}}$

Doen, oefen met een medeleerling.

Doen.

Oefen met een medeleerling.

`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.

`(1 + text(i))^5 = (sqrt(2)*text(e)^(1/4 pi text(i)))^5 = 4sqrt(2)*text(e)^(5/4 pi text(i))= 4sqrt(2)(cos(5/4 pi) + text(i)sin(5/4 pi)) = text(-)4-4text(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.

`(2 + text(i))^2 = 4 + 4text(i) + (text(i))^2 = 3 + 4text(i)`

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

Je krijgt `sqrt(3+4text(i)) = (3+4text(i))^(0,5) ~~ (5*text(e)^(0,93text(i)+k*2pi))^(0,5) = 5^(0,5)*text(e)^(0,5*0,93text(i)+k*pi)` en dus krijg je `z_1 = sqrt(5)text(e)^(0,46text(i)) vv z_1 = sqrt(5)text(e)^(0,46text(i)+pi)` .

Dus krijg je behalve `z_1 = 2 + text(i)` ook als mogelijkheid `z_1 = text(-)2 - text(i)` .

`sqrt(text(-)3) * sqrt(text(-)12) = text(i)sqrt(3) * text(i)sqrt(12) = text(i)^2*sqrt(3*12) = text(-)sqrt(36) = text(-)6` .

Maar `sqrt(text(-)3 * text(-)12) = sqrt(36) = 6` .

Ook hier ontstaat het probleem doordat wortels uit complexe getallen niet éénduidig zijn. Bijvoorbeeld `sqrt(text(-)3)` is niet alleen gelijk aan `text(i)sqrt(3)` , maar ook aan `text(-i)sqrt(3)` .

De boodschap is om met wortels uit complexe getallen voorzichtig te zijn.

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

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

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

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

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

$z=2\sqrt{2}\cdot {\text{e}}^{\text{-}\mathrm{0,75}\pi \text{i}}$

`2 - 2text(i) = sqrt(8)text(e)^(text(-)1/4 pi text(i))` en dus `(2 - 2text(i))^5 = sqrt(8)^5 * text(e)^(text(-)5/4 pi text(i))` .

Dus `(2 - 2text(i))^5 = text(-)128 + 128text(i)` .

`(2 - 3text(i))^5 = (sqrt(13)text(e)^(text(-)0,98... text(i)))^5 = (sqrt(13))^5 text(e)^(text(-)4,91... text(i))` .

Dus `(2 - 3text(i))^5 =122+597text(i)` (niet afronden tussentijds!).

Vervang `varphi` door `text(-)varphi` en gebruik `sin(text(-)varphi) = text(-)sin(varphi)` en `cos(text(-)varphi) = cos(varphi)` .

`text(e)^(text(i)varphi)+text(e)^(text(-)text(i)varphi)=cos(varphi)+text(i)sin(varphi) + cos(varphi)-text(i)sin(varphi)=2cos(varphi)`

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

`text(i) = 1*text(e)^(1/2 pi text(i))` en `3-5text(i) = sqrt(34)*text(e)^(text(-)1,030text(i))`

`z = sqrt(34)*text(e)^((1/2 pi - 1,030)text(i)) = 5+3text(i)` (niet tussentijds afronden).

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

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

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