If the line $z\overline\alpha + \overline z\alpha + i\beta = 0$ makes an angle of $45^\circ$ with the real axis, the value of $(1+i)(-\frac {2\alpha} {\overline\alpha})$ is
a) $2\sqrt2$
b) $2\sqrt2 i$
c) $2(1-i)$
d) $-2(1+i)$
My attempt:
The complex slope of the line is $-\frac{\alpha}{\overline \alpha}$. This is equal to $e^{i\frac \pi 4}$ as it makes an angle of $45^\circ$. Substituting this in the expression gives
$$2(1+i)e^{i\frac \pi 4} = 2\sqrt2 e^{i\frac \pi 2} = 2\sqrt{2} i$$
which corresponds to option b. However, my textbook gives c as the answer. Please verify if my solution is correct or if it is an error in the textbook.
Best Answer
The error in your solution is that, if the line makes an angle of $45°$ with the real axis doesn't imply that the complex slope is $e^{\frac{iπ}{4}}$.
Let's keep it simple. Let $z=x+iy$. So, given equation can be written as, $\begin{align} \bar{a}(x+iy)+a(x-iy)+ib=0\\ x(\bar{a}+a)+yi(\bar{a}-a)+ib=0 \end{align}$
Now, slope of this line is $\begin{align} -\frac{c.e\,of\,x}{c.e\,of\,y}&=-\frac{\bar{a}+a}{i(\bar{a}-a)}=tan\frac{π}{4}\\ \therefore -\frac{a}{\bar a}&=i \end{align}$
After substituting in the required expression, we get, $$2(i-1)$$ According to me, none of the options is matching or there may be correction in option $C$.