Keegan Stevens

2023-03-25

How to express the complex number in trigonometric form: 5-5i?

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$5\sqrt{2}\left(\mathrm{cos}\left(-\frac{\pi }{4}\right)+i\mathrm{sin}\left(-\frac{\pi }{4}\right)\right)$
Explanation:
To convert from $z=x+iy$ form to $z=r\left(\mathrm{cos}\theta +i\mathrm{sin}\theta \right)$ form, you need to find $r$ and $\theta$.
$r=\sqrt{{x}^{2}+{y}^{2}}$ and $\mathrm{tan}\theta =\frac{y}{x}$
So, $r=\sqrt{{5}^{2}+{\left(-5\right)}^{2}}=\sqrt{50}=5\sqrt{2}$
$\mathrm{tan}\theta =\frac{5}{-5}⇒\mathrm{tan}\theta =-1⇒\theta =-\frac{\pi }{4}$
$\therefore 5-5i=5\sqrt{2}\left(\mathrm{cos}\left(-\frac{\pi }{4}\right)+i\mathrm{sin}\left(-\frac{\pi }{4}\right)\right)$

Kieran Orozco

Trigonometric form: $7.07\left(\mathrm{cos}45-i\mathrm{sin}45\right)$ Explanation: $Z=a+ib=5-5i$. Modulus:$|Z|=\sqrt{{a}^{2}+{b}^{2}}$
Modulus:$|Z|=\sqrt{{5}^{2}+{\left(-5\right)}^{2}}$
$=\sqrt{50}\approx 7.07$ Argument: $\mathrm{tan}\alpha =|\frac{b}{a}|=|\frac{5}{-5}|=1$
$\therefore \alpha ={\mathrm{tan}}^{-1}\left(1\right)={45}^{0};Z$ lies on fourth
quadrant.$\therefore \theta =360-\alpha ={315}^{0}\phantom{\rule{1ex}{0ex}}\text{or}\phantom{\rule{1ex}{0ex}}{\left(-45\right)}^{0}$
as represented in trigonometric form as
$Z=|Z|\left(\mathrm{cos}\theta +i\mathrm{sin}\theta \right)$
$\therefore Z=7.07\left(\mathrm{cos}315+i\mathrm{sin}315\right)$ or
$Z=7.07\left(\mathrm{cos}45-i\mathrm{sin}45\right)$

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