How do you write the complex number in standard form #3/2(cos300+isin300)#?

Question

Lily Anderson · Accepted Answer

#3/4 - (3sqrt3)/4i#

I'll assume the trigonometric arguments are in degrees.

We're asked to find the standard form of

#3/2(cos[300^"o"] + isin[300^"o"])#

Well, #cos[300^"o"] = cos[-60^"o"] = color(red)(1/2#

And #sin[300^"o"] = sin[-60^"o"] = color(green)(-sqrt3/2#

So we have

#3/2(1/2 - (isqrt3)/2) = color(blue)(ulbar(|stackrel(" ")(" "3/4 - (3sqrt3)/4i" ")|)#

Anna Allen · Answer

undefined To write the complex number $ \frac{3}{2} (\cos 300^\circ + i \sin 300^\circ) $ in standard form, you can use Euler's formula, which states that $ e^{i\theta} = \cos \theta + i \sin \theta $.

First, convert $ 300^\circ $ to radians by using the conversion factor $ \frac{\pi}{180} $ since $ 1^\circ = \frac{\pi}{180} $ radians.

$ 300^\circ \times \frac{\pi}{180} = \frac{5\pi}{6} $ radians.

Now, substitute $ \frac{5\pi}{6} $ into Euler's formula:

$ e^{i \frac{5\pi}{6}} = \cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} $

$ = -\frac{\sqrt{3}}{2} + i \left(-\frac{1}{2}\right) $

Now, multiply $ \frac{3}{2} $ by this result to get the complex number in standard form:

$ \frac{3}{2} \times \left( -\frac{\sqrt{3}}{2} + i \left(-\frac{1}{2}\right) \right) $

$ = -\frac{3\sqrt{3}}{4} - \frac{3i}{4} $

So, the complex number $ \frac{3}{2} (\cos 300^\circ + i \sin 300^\circ) $ in standard form is $ -\frac{3\sqrt{3}}{4} - \frac{3i}{4} $.