Trig/Complex Numbers Problem in Riley Math Methods

Question

I'm having trouble with a solution to a problem in Riley's Math Methods book. I've presented the problem and solution in the two pictures below (question b) only), but essentially I am wondering why, in the case where the two are out of phase by $\pi$, we can then easily conclude that $c_ib_r = c_rb_i$. My thinking is as follows. If the two are $\pi$ radians out of phase, then the angles (phases) of B and C are related by $$ \pm\pi = tan^{-1}(b_i/b_r) - tan^{-1}(c_i/c_r) $$ but from there I cannot see how we can conclude that $b_i/b_r = c_i/c_r$ and thus that $c_ib_r = c_rb_i$. In the case where their phase difference is $0$ this is easily seen.

Question b) only

Solution

Answer 1

If $b$ and $c$ are $\pi$ out of phase, then $\Im bc^*=-b_rc_i+b_ic_r=0$. I guess the easiest way to see that is that if $b=r_be^{i\theta}$ then $c=r_ce^{i(\theta+\pi)}=-r_ce^{i\theta}$ so $bc^*=-r_br_c\in\mathbb{R}$. But what I would have done is to say that $$z=a+bt+ct^2=a-\frac{b^2}{4c^2}c+\left(t+\frac b{2c}\right)^2c$$ That way you can see that it's a ray starting at $$z=a-\frac{b^2}{4c^2}c$$ And going in the direction of $c$. Also from the ray formulation you can get $$x=a_r-\frac{b^2}{4c^2}c_r+\left(t-\frac b{2c}\right)^2c_r$$ $$y=a_i-\frac{b^2}{4c^2}c_i+\left(t-\frac b{2c}\right)^2c_i$$ Multiplying the first equation above by $c_i$ and the second by $c_r$ and subtracting you get $$c_ix-c_ry=a_rc_i-a_ic_r$$ In the first question make sure you exclude $z=c-\rho$!

Answer 2

Let $\dfrac {b_i}{c_i} = V$ and $\dfrac {b_r}{c_r} = W$. If $\tan W = \tan V$ we get the solution (after taking the arctangent of both sides) $$V = \pi k + W$$

If we let $k=0$ $$V = W \rightarrow \dfrac {b_i}{c_i} = \dfrac {b_r}{c_r}$$ and hence $${b_i}{c_r} = {b_r}{c_i} \implies {b_i}{c_r} - {b_r}{c_i} = 0$$ and thus $t$ vanishes (as $t \cdot 0 = 0$).