NT Warm-up

Algebra Level 2

Express ( 1 2 + 3 2 i ) 2015 \left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i\right)^{2015} in the form a + b i a+bi .

1 2 3 2 i \frac{1}{2} - \frac{\sqrt{3}}{2}i 1 4 3 2 i \frac{1}{4} - \frac{\sqrt{3}}{2}i 1 2 + 3 2 i \frac{1}{2} + \frac{\sqrt{3}}{2}i 1 4 + 3 2 i \frac{1}{4} + \frac{\sqrt{3}}{2}i

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Chew-Seong Cheong
Aug 23, 2018

z = ( 1 2 + 3 2 i ) 2015 = ( cos π 3 + i sin π 3 ) 2015 By Euler formula = ( e π 3 i ) 2015 = e 671 2 3 i Since e 2 π = 1 = e 5 3 π i = cos 5 3 π + i sin 5 3 π = 1 2 3 2 i \begin{aligned} z & = \left(\frac 12 + \frac {\sqrt 3}2i \right)^{2015} \\ & = \left(\color{#3D99F6} \cos \frac \pi 3 + i\sin \frac \pi 3 \right)^{2015} & \small \color{#3D99F6} \text{By Euler formula} \\ & = \left(\color{#3D99F6} e^{\frac \pi 3 i} \right)^{2015} = e^{671\frac 23i} & \small \color{#3D99F6} \text{Since }e^{2\pi} = 1 \\ & = e^{\frac 53\pi i} \\ & = \cos \frac 53 \pi + i \sin \frac 53 \pi \\ & = \boxed {\dfrac 12 - \dfrac {\sqrt 3}2 i} \end{aligned}

X X
Aug 23, 2018

( 1 2 + 3 2 i ) 2015 \left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i\right)^{2015}

= ( cos π 3 + i sin π 3 ) 2015 =\left(\cos\dfrac{\pi}3+i\sin\dfrac{\pi}3\right)^{2015}

= cos 2015 π 3 + i sin 2015 π 3 =\cos\dfrac{2015\pi}3+i\sin\dfrac{2015\pi}3

= 1 2 3 2 i =\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...