# Write the complex number in trigonometric form r(cos theta + i sin theta),with theta in the interval [0^@, 360^@]5sqrt3 + 5i

Write the complex number in trigonometric form $r\left(\mathrm{cos}\theta +i\mathrm{sin}\theta \right)$,with theta in the interval $\left[{0}^{\circ },{360}^{\circ }\right]$
$5\sqrt{3}+5i$

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pierretteA
The trigonometric form of the complex number is $r\left(\mathrm{cos}\theta +i\mathrm{sin}\theta \right).$
Here, r is the modulus of the complex number and theta is called the argument of the complex number.
The formula for the modulus of any complex number a+bi is defined as, $r=\sqrt{{a}^{2}+{b}^{2}}$
Substitute $a=5\sqrt{3},b=5$ in the formula for modulus.
$r=\sqrt{{\left(5\sqrt{3}\right)}^{2}+{5}^{2}=\sqrt{75+25}=\sqrt{100}=10}$
Since r is the modulus, it is always positive. So, r= 10.
The formula for the argument of the complex number is defined as, $\theta ={\mathrm{tan}}^{-1}\left(\frac{b}{a}\right)$
Substitute $a=5\sqrt{3},b=5$ in the formula for argument
$\theta ={\mathrm{tan}}^{-1}\left(\frac{5}{5\sqrt{3}}\right)={\mathrm{tan}}^{-1}\left(\frac{1}{\sqrt{3}}\right)$
The value of tan is $\frac{1}{\sqrt{3}}$ for $\frac{\pi }{6}$
So, $\theta =\frac{\pi }{6}$
So, the trigonometric form of the given complex number is $10\left(\mathrm{cos}\left(\frac{\pi }{6}\right)+i\mathrm{sin}\left(\frac{\pi }{6}\right)$