JEE Main 2024 — Mathematics Algebra

Given,

$${|1-i|}^{x}={2}^{x}$$∣1−i∣x=2x

$$\Rightarrow {(\sqrt{2})}^{x}={2}^{x}$$⇒(2​)x=2x

$$\Rightarrow x=0$$⇒x=0

Hence, the number of solutions will be one, so $\alpha =1$α=1

Now, solving

$$z=\frac{\pi }{4}{(1+i)}^{4}(\frac{1-\sqrt{\pi }\cdot i}{\sqrt{\pi }+i}+\frac{\sqrt{\pi }-i}{1+\sqrt{\pi }\cdot i})$$z=4π​(1+i)4(π​+i1−π​⋅i​+1+π​⋅iπ​−i​)

$$\Rightarrow z=\frac{\pi }{4}{(1+i)}^{4}[\frac{\sqrt{\pi }-\pi i-i-\sqrt{\pi }}{\pi +1}+\frac{\sqrt{\pi }-i-\pi i-\sqrt{\pi }}{1+\pi }]$$⇒z=4π​(1+i)4[π+1π​−πi−i−π​​+1+ππ​−i−πi−π​​]

$$\Rightarrow z=\frac{\pi }{4}{(1+i)}^{4}[\frac{-2\pi i-2i}{\pi +1}]$$⇒z=4π​(1+i)4[π+1−2πi−2i​]

$$\Rightarrow z=\frac{\pi i}{2}{(1+i)}^{4}$$⇒z=2πi​(1+i)4

$$\Rightarrow z=-\frac{\pi i}{2}(1+4i+6{i}^{2}+4{i}^{3}+1)$$⇒z=−2πi​(1+4i+6i2+4i3+1)

$$\Rightarrow z=2\pi i$$⇒z=2πi

So, $arg(z)=\frac{\pi }{2}$arg(z)=2π​

$$\Rightarrow \beta =\frac{|z|}{arg(z)}=\frac{2\pi }{\frac{\pi }{2}}=4$$⇒β=arg(z)∣z∣​=2π​2π​=4

Hence, the distance from $(\alpha ,\beta )\equiv (1,4)$(α,β)≡(1,4) to $4x-3y=7$4x−3y=7 will be $|\frac{4-3\times 4-7}{\sqrt{{4}^{2}+{3}^{2}}}|=\frac{15}{5}=3$∣42+32​4−3×4−7​∣=515​=3