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Mathematics 21 Online
OpenStudy (anonymous):

help :(

OpenStudy (anonymous):

jhonyy9 (jhonyy9):

- so for point b. i think 45 degree

OpenStudy (experimentx):

modulus = sqrt(3^2+3^2) = 3 sqrt(2) angle = 135 => arctan(-3/3)

OpenStudy (anonymous):

absolute values is the lenght which you get by pythagoras. for a complex number \(a+bi\) we have \(|a+bi|=\sqrt{a^b+b^2}\) in your case it is \[|-3+3i|=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}\]

OpenStudy (anonymous):

and the angle would be -45?

OpenStudy (anonymous):

no and this is where you have to be careful!

OpenStudy (anonymous):

it is true that \(\tan(\frac{b}{a})=\theta\) but it is not necessarily true that \(\theta=\tan^{-1}(\frac{b}{a})\)

OpenStudy (anonymous):

you have to consider what quadrant you are in. arctan only returns values from -90 to 90, but from your picture you see that you are in quadrant II not in I or IV so you have to think what angle that is, it is not -45 but rather 135

sam (.sam.):

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