[tex] | a + bi | = \sqrt{(a)^2+(b)^2} [/tex](adsbygoogle = window.adsbygoogle || []).push({});

[tex] \theta = tan^{-1}(\frac{b}{a}) [/tex]

[tex] |(a + bi)^n| = (\sqrt{(a)^2+(b)^2})^n[/tex]

[tex] \theta = n(tan^{-1}(\frac{b}{a}))[/tex]

[tex] (a+bi) = (\sqrt{(a)^2+(b)^2})^n(cos(n(tan^{-1}(\frac{b}{a})) + isin(n(tan^{-1}(\frac{b}{a}))))[/tex]

Why is [tex]\theta = n(tan^{-1}(\frac{b}{a}) [/tex] ? For example when I have

[tex]5 + i[/tex]

[tex]\theta = tan^{-1}(\frac{1}{5}) = 0.1974...[/tex]

[tex](5+i)^2 = 25 + 10i -1[/tex]

[tex](5+i)^2 = 24 + 10i[/tex]

[tex]\alpha = tan^{-1}(\frac{10}{24}) = 0.3948...[/tex]

[tex]2(0.1974) =0.3948[/tex]

Why is it that the exponent of the vector can be used to get the angle of the resultant by simply multiplying it with the tan function? Also in the first part of that:

[tex] (\sqrt{(a)^2+(b)^2})^n(cos(n(tan^{-1}(\frac{b}{a})) + isin(n(tan^{-1}(\frac{b}{a}))))[/tex]

Just to make sure, it is only

[tex] isin(n(tan^{-1}(\frac{b}{a}))))[/tex]

and not simply

[tex] (\sqrt{(a)^2+(b)^2})^n(cos(n(tan^{-1}(\frac{b}{a})) + sin(n(tan^{-1}(\frac{b}{a}))))[/tex]

because one of the components is complex?

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# Trig question

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