If $$\alpha$$ denotes the number of solutions of $$|1 - i|^x = 2^x$$ and $$\beta = \frac{|z|}{\arg(z)}$$, where $$z = \frac{\pi}{4}(1+i)^4\left(\frac{1-\sqrt{\pi}\cdot i}{\sqrt{\pi}+i} + \frac{\sqrt{\pi}-i}{1+\sqrt{\pi}\cdot i}\right)$$, $$i = \sqrt{-1}$$, then the distance of the point $$(\alpha, \beta)$$ from the line $$4x - 3y = 7$$ is ______

We first determine $$\alpha$$ from the equation $$|1-i|^{\,x}=2^{x}$$.

The modulus of $$1-i$$ is $$|1-i|=\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}$$. Hence the equation becomes$$(\sqrt{2})^{\,x}=2^{x}\quad -(1)$$

Writing both sides with base $$2$$ gives $$2^{x/2}=2^{x}$$. Equating the exponents,$$\frac{x}{2}=x\;\Longrightarrow\;x=0$$

Therefore there is exactly one real solution, so $$\alpha=1$$.

Next we evaluate $$\beta=\dfrac{|z|}{\arg(z)}$$ for $$z=\dfrac{\pi}{4}(1+i)^{4}\Bigl(\dfrac{1-\sqrt{\pi}\,i}{\sqrt{\pi}+i}+\dfrac{\sqrt{\pi}-i}{1+\sqrt{\pi}\,i}\Bigr)$$. Let $$s=\sqrt{\pi}$$ for brevity.

1. Compute $$(1+i)^{4}$$: $$1+i=\sqrt{2}\,e^{\,i\pi/4}\Longrightarrow(1+i)^{4}=(\sqrt{2})^{4}e^{\,i\pi}=4(-1)=-4$$

2. Evaluate the bracketed sum $$S$$: $$S=\dfrac{1-si}{s+i}+\dfrac{s-i}{1+si}$$

Case 1: $$A=\dfrac{1-si}{s+i}$$
Multiply numerator and denominator by $$s-i$$: $$A=\dfrac{(1-si)(s-i)}{(s+i)(s-i)}=\dfrac{(s-i)-s^{2}i-s}{s^{2}+1}=\dfrac{-(1+s^{2})\,i}{s^{2}+1}=-i$$

Case 2: $$B=\dfrac{s-i}{1+si}$$
Multiply numerator and denominator by $$1-si$$: $$B=\dfrac{(s-i)(1-si)}{(1+si)(1-si)}=\dfrac{s-s^{2}i-i-s}{1+s^{2}}=\dfrac{-(1+s^{2})\,i}{1+s^{2}}=-i$$

Hence $$S=A+B=-i-i=-2i$$.

3. Combine everything to get $$z$$: $$z=\dfrac{\pi}{4}\cdot(-4)\cdot(-2i)=2\pi i$$

Modulus: $$|z|=|2\pi i|=2\pi$$
Argument: since $$z$$ lies on the positive imaginary axis, $$\arg(z)=\dfrac{\pi}{2}$$.

Therefore $$\beta=\dfrac{|z|}{\arg(z)}=\dfrac{2\pi}{\pi/2}=4$$.

The required point is $$(\alpha,\beta)=(1,4)$$.

Distance of $$(1,4)$$ from the line $$4x-3y=7$$ is$$\displaystyle d=\frac{|4(1)-3(4)-7|}{\sqrt{4^{2}+(-3)^{2}}}=\frac{|4-12-7|}{5}=\frac{15}{5}=3$$

Hence the required distance is $$3$$.