Bepaal modulus en argument
- \(10+2i\)
- \(6+6i\)
- \(-2-5i\)
- \(4-4i\)
- \(-10-5i\)
- \(-2+i\)
- \(1\)
- \(-7-8i\)
- \(5-10i\)
- \(9-8i\)
- \(-3+7i\)
- \(2+7i\)
Bepaal modulus en argument
Verbetersleutel
- \(10+2i\\ r = \sqrt{10^2+2^2} = \sqrt{104} \\ \alpha = tan^{-1}(\frac{2}{10}) \Leftrightarrow \alpha =11^\circ 18' 35{,}8"\text{ of } \alpha = 191^\circ 18' 35{,}8"\\10+2i\text{ ligt in kwadrant }1, \alpha \text{ ligt dus tussen }0^\circ \text{ en }90^\circ\\ \alpha = 11^\circ 18' 35{,}8"\)
- \(6+6i\\ r = \sqrt{6^2+6^2} = \sqrt{72} \\ \alpha = tan^{-1}(\frac{6}{6}) \Leftrightarrow \alpha =45^\circ \text{ of } \alpha = 225^\circ \\6+6i\text{ ligt in kwadrant }1, \alpha \text{ ligt dus tussen }0^\circ \text{ en }90^\circ\\ \alpha = 45^\circ \)
- \(-2-5i\\ r = \sqrt{(-2)^2+(-5)^2} = \sqrt{29} \\ \alpha = tan^{-1}(\frac{-5}{-2}) \Leftrightarrow \alpha =68^\circ 11' 54{,}9"\text{ of } \alpha = 248^\circ 11' 54{,}9"\\-2-5i\text{ ligt in kwadrant }3, \alpha \text{ ligt dus tussen }180^\circ \text{ en }270^\circ\\ \alpha = 248^\circ 11' 54{,}9"\)
- \(4-4i\\ r = \sqrt{4^2+(-4)^2} = \sqrt{32} \\ \alpha = tan^{-1}(\frac{-4}{4}) \Leftrightarrow \alpha =135^\circ \text{ of } \alpha = 315^\circ \\4-4i\text{ ligt in kwadrant }4, \alpha \text{ ligt dus tussen }270^\circ \text{ en }360^\circ\\ \alpha = 315^\circ \)
- \(-10-5i\\ r = \sqrt{(-10)^2+(-5)^2} = \sqrt{125} \\ \alpha = tan^{-1}(\frac{-5}{-10}) \Leftrightarrow \alpha =26^\circ 33' 54{,}2"\text{ of } \alpha = 206^\circ 33' 54{,}2"\\-10-5i\text{ ligt in kwadrant }3, \alpha \text{ ligt dus tussen }180^\circ \text{ en }270^\circ\\ \alpha = 206^\circ 33' 54{,}2"\)
- \(-2+i\\ r = \sqrt{(-2)^2+1^2} = \sqrt{5} \\ \alpha = tan^{-1}(\frac{1}{-2}) \Leftrightarrow \alpha =153^\circ 26' 5{,}8"\text{ of } \alpha = 333^\circ 26' 5{,}8"\\-2+i\text{ ligt in kwadrant }2, \alpha \text{ ligt dus tussen }90^\circ \text{ en }180^\circ\\ \alpha = 153^\circ 26' 5{,}8"\)
- \(1\\ \text{ Dit complex getal ligt op het positief gedeelte van de x-as. We hebben geen berekeningen nodig om r of } \alpha \text{ te berekenen.} \\\text{r = }1\\\alpha = 0 ^\circ \\\)
- \(-7-8i\\ r = \sqrt{(-7)^2+(-8)^2} = \sqrt{113} \\ \alpha = tan^{-1}(\frac{-8}{-7}) \Leftrightarrow \alpha =48^\circ 48' 50{,}7"\text{ of } \alpha = 228^\circ 48' 50{,}7"\\-7-8i\text{ ligt in kwadrant }3, \alpha \text{ ligt dus tussen }180^\circ \text{ en }270^\circ\\ \alpha = 228^\circ 48' 50{,}7"\)
- \(5-10i\\ r = \sqrt{5^2+(-10)^2} = \sqrt{125} \\ \alpha = tan^{-1}(\frac{-10}{5}) \Leftrightarrow \alpha =116^\circ 33' 54{,}2"\text{ of } \alpha = 296^\circ 33' 54{,}2"\\5-10i\text{ ligt in kwadrant }4, \alpha \text{ ligt dus tussen }270^\circ \text{ en }360^\circ\\ \alpha = 296^\circ 33' 54{,}2"\)
- \(9-8i\\ r = \sqrt{9^2+(-8)^2} = \sqrt{145} \\ \alpha = tan^{-1}(\frac{-8}{9}) \Leftrightarrow \alpha =138^\circ 21' 59{,}3"\text{ of } \alpha = 318^\circ 21' 59{,}3"\\9-8i\text{ ligt in kwadrant }4, \alpha \text{ ligt dus tussen }270^\circ \text{ en }360^\circ\\ \alpha = 318^\circ 21' 59{,}3"\)
- \(-3+7i\\ r = \sqrt{(-3)^2+7^2} = \sqrt{58} \\ \alpha = tan^{-1}(\frac{7}{-3}) \Leftrightarrow \alpha =113^\circ 11' 54{,}9"\text{ of } \alpha = 293^\circ 11' 54{,}9"\\-3+7i\text{ ligt in kwadrant }2, \alpha \text{ ligt dus tussen }90^\circ \text{ en }180^\circ\\ \alpha = 113^\circ 11' 54{,}9"\)
- \(2+7i\\ r = \sqrt{2^2+7^2} = \sqrt{53} \\ \alpha = tan^{-1}(\frac{7}{2}) \Leftrightarrow \alpha =74^\circ 3' 16{,}6"\text{ of } \alpha = 254^\circ 3' 16{,}6"\\2+7i\text{ ligt in kwadrant }1, \alpha \text{ ligt dus tussen }0^\circ \text{ en }90^\circ\\ \alpha = 74^\circ 3' 16{,}6"\)