Complex or Simple

Algebra Level 3

If a = e i α a = e^{i\alpha} , b = e i β b = e^{i\beta} and c = e i γ c = e^{i\gamma} and a b + b c + c a = 1 \dfrac ab + \dfrac bc + \dfrac ca = 1 , find c y c cos ( α β ) \displaystyle \sum_{cyc} \cos (\alpha - \beta) .

3 2 \frac 32 3 2 -\frac 32 0 1

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.

1 solution

Chew-Seong Cheong
Mar 15, 2018

a b + b c + c a = 1 e i α e i β + e i β e i γ + e i γ e i α = 1 e i ( α β ) + e i ( β γ ) + e i ( γ α ) = 1 Using Euler’s formula: e i θ = cos θ + i sin θ cos ( α β ) + i sin ( α β ) + cos ( β γ ) + i sin ( β γ ) + cos ( γ α ) + i sin ( γ α ) = 1 c y c cos ( α β ) + i c y c sin ( α β ) = 1 + i 0 Equating the real part c y c cos ( α β ) = 1 \begin{aligned} \frac ab + \frac bc + \frac ca & = 1 \\ \frac {e^{i\alpha}}{e^{i\beta}} + \frac {e^{i\beta}}{e^{i\gamma}} + \frac {e^{i\gamma}}{e^{i\alpha}} & = 1 \\ e^{i(\alpha - \beta)} + e^{i(\beta - \gamma)} + e^{i(\gamma - \alpha)} & = 1 & \small \color{#3D99F6} \text{Using Euler's formula: }e^{i\theta} = \cos \theta + i\sin \theta \\ \cos (\alpha - \beta) + i\sin (\alpha - \beta) + \cos (\beta - \gamma) + i \sin (\beta - \gamma) + \cos (\gamma - \alpha) + i\sin (\gamma - \alpha) & = 1 \\ \sum_{cyc} \cos (\alpha - \beta) + i \sum_{cyc} \sin (\alpha - \beta) & = 1 + i0 & \small \color{#3D99F6} \text{Equating the real part} \\ \implies \sum_{cyc} \cos (\alpha - \beta) & = \boxed{1} \end{aligned}

How c y c sin ( α β ) = 0 \sum_{cyc} \sin (\alpha - \beta)=0 ?

Sahil Silare - 3 years, 2 months ago

Log in to reply

2 Complex Numbers are equal if their real and imaginary parts are equal.

Moulik Bhattacharya - 3 years, 2 months ago

Log in to reply

What are you tryin to say? Please elaborate.

Sahil Silare - 3 years, 2 months ago

Log in to reply

@Sahil Silare Let { z 1 = x 1 + i y 1 z 2 = x 2 + i y 2 \begin{cases} z_1 = x_1 + i y_1 \\ z_2 = x_2 + i y_2 \end{cases} , z 1 = z 2 z_1 = z_2 if and only if x 1 = x 2 x_1 = x_2 or ( z 1 ) = ( z 2 ) \Re (z_1) = \Re (z_2) and y 1 = y 2 y_1 = y_2 or ( z 1 ) = ( z 2 ) \Im (z_1) = \Im (z_2) .

Chew-Seong Cheong - 3 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...