Bepaal modulus en argument
- \(1+i\)
- \(8-2i\)
- \(-7-8i\)
- \(5-6i\)
- \(3-i\)
- \(-4-i\)
- \(-6i\)
- \(10+9i\)
- \(-1-4i\)
- \(-5+4i\)
- \(-3+10i\)
- \(3-10i\)
Bepaal modulus en argument
Verbetersleutel
- \(1+i\\ r = \sqrt{1^2+1^2} = \sqrt{2} \\ \alpha = tan^{-1}(\frac{1}{1}) \Leftrightarrow \alpha =45^\circ \text{ of } \alpha = 225^\circ \\1+i\text{ ligt in kwadrant }1, \alpha \text{ ligt dus tussen }0^\circ \text{ en }90^\circ\\ \alpha = 45^\circ \)
- \(8-2i\\ r = \sqrt{8^2+(-2)^2} = \sqrt{68} \\ \alpha = tan^{-1}(\frac{-2}{8}) \Leftrightarrow \alpha =165^\circ 57' 49{,}5"\text{ of } \alpha = 345^\circ 57' 49{,}5"\\8-2i\text{ ligt in kwadrant }4, \alpha \text{ ligt dus tussen }270^\circ \text{ en }360^\circ\\ \alpha = 345^\circ 57' 49{,}5"\)
- \(-7-8i\\ r = \sqrt{(-7)^2+(-8)^2} = \sqrt{113} \\ \alpha = tan^{-1}(\frac{-8}{-7}) \Leftrightarrow \alpha =48^\circ 48' 50{,}7"\text{ of } \alpha = 228^\circ 48' 50{,}7"\\-7-8i\text{ ligt in kwadrant }3, \alpha \text{ ligt dus tussen }180^\circ \text{ en }270^\circ\\ \alpha = 228^\circ 48' 50{,}7"\)
- \(5-6i\\ r = \sqrt{5^2+(-6)^2} = \sqrt{61} \\ \alpha = tan^{-1}(\frac{-6}{5}) \Leftrightarrow \alpha =129^\circ 48' 20{,}1"\text{ of } \alpha = 309^\circ 48' 20{,}1"\\5-6i\text{ ligt in kwadrant }4, \alpha \text{ ligt dus tussen }270^\circ \text{ en }360^\circ\\ \alpha = 309^\circ 48' 20{,}1"\)
- \(3-i\\ r = \sqrt{3^2+(-1)^2} = \sqrt{10} \\ \alpha = tan^{-1}(\frac{-1}{3}) \Leftrightarrow \alpha =161^\circ 33' 54{,}2"\text{ of } \alpha = 341^\circ 33' 54{,}2"\\3-i\text{ ligt in kwadrant }4, \alpha \text{ ligt dus tussen }270^\circ \text{ en }360^\circ\\ \alpha = 341^\circ 33' 54{,}2"\)
- \(-4-i\\ r = \sqrt{(-4)^2+(-1)^2} = \sqrt{17} \\ \alpha = tan^{-1}(\frac{-1}{-4}) \Leftrightarrow \alpha =14^\circ 2' 10{,}5"\text{ of } \alpha = 194^\circ 2' 10{,}5"\\-4-i\text{ ligt in kwadrant }3, \alpha \text{ ligt dus tussen }180^\circ \text{ en }270^\circ\\ \alpha = 194^\circ 2' 10{,}5"\)
- \(-6i\\ \text{ Dit complex getal ligt op het negatief gedeelte van de y-as. We hebben geen berekeningen nodig om r of } \alpha \text{ te berekenen.} \\\text{r = }6\\\alpha = 270 ^\circ \\\)
- \(10+9i\\ r = \sqrt{10^2+9^2} = \sqrt{181} \\ \alpha = tan^{-1}(\frac{9}{10}) \Leftrightarrow \alpha =41^\circ 59' 14"\text{ of } \alpha = 221^\circ 59' 14"\\10+9i\text{ ligt in kwadrant }1, \alpha \text{ ligt dus tussen }0^\circ \text{ en }90^\circ\\ \alpha = 41^\circ 59' 14"\)
- \(-1-4i\\ r = \sqrt{(-1)^2+(-4)^2} = \sqrt{17} \\ \alpha = tan^{-1}(\frac{-4}{-1}) \Leftrightarrow \alpha =75^\circ 57' 49{,}5"\text{ of } \alpha = 255^\circ 57' 49{,}5"\\-1-4i\text{ ligt in kwadrant }3, \alpha \text{ ligt dus tussen }180^\circ \text{ en }270^\circ\\ \alpha = 255^\circ 57' 49{,}5"\)
- \(-5+4i\\ r = \sqrt{(-5)^2+4^2} = \sqrt{41} \\ \alpha = tan^{-1}(\frac{4}{-5}) \Leftrightarrow \alpha =141^\circ 20' 24{,}7"\text{ of } \alpha = 321^\circ 20' 24{,}7"\\-5+4i\text{ ligt in kwadrant }2, \alpha \text{ ligt dus tussen }90^\circ \text{ en }180^\circ\\ \alpha = 141^\circ 20' 24{,}7"\)
- \(-3+10i\\ r = \sqrt{(-3)^2+10^2} = \sqrt{109} \\ \alpha = tan^{-1}(\frac{10}{-3}) \Leftrightarrow \alpha =106^\circ 41' 57{,}3"\text{ of } \alpha = 286^\circ 41' 57{,}3"\\-3+10i\text{ ligt in kwadrant }2, \alpha \text{ ligt dus tussen }90^\circ \text{ en }180^\circ\\ \alpha = 106^\circ 41' 57{,}3"\)
- \(3-10i\\ r = \sqrt{3^2+(-10)^2} = \sqrt{109} \\ \alpha = tan^{-1}(\frac{-10}{3}) \Leftrightarrow \alpha =106^\circ 41' 57{,}3"\text{ of } \alpha = 286^\circ 41' 57{,}3"\\3-10i\text{ ligt in kwadrant }4, \alpha \text{ ligt dus tussen }270^\circ \text{ en }360^\circ\\ \alpha = 286^\circ 41' 57{,}3"\)