Let z be a complex number such that the real part of z - 2i / z + 2i is zero. Then, the maximum value

Question

Let z be a complex number such that the real part of \(\frac{z-2 i}{z+2 i}\) is zero. Then, the maximum value of \(|z-(6+8 i)|\) is equal to : 

(1) 12

(2) \( \infty\)

(3) 10

(4) 8

Answer 1

Correct option is : (1) 12 

\( \operatorname{Re}\left(\frac{z-2 i}{z+2 i}\right)=0 \) 

Let \(\mathrm{z}=\mathrm{x}+\mathrm{iy}\) 

\( \frac{x+i y-2 i}{x+i y+2 i} \Rightarrow \frac{x+i(y-2)}{x+i(y+2)} \) 

Rationalizing 

\( \begin{aligned} \Rightarrow \frac{x+i(y-2)}{x+i(y+2)} \times \frac{x-i(y+2)}{x-i(y+2)} \end{aligned} \) 

\( \begin{aligned} \Rightarrow \frac{x^2+\left(y^2-4\right)+i(x y-2 x-x y-2 x)}{x^2+(y+2)^2} \end{aligned} \)

Now, \(\operatorname{Re}\left(\frac{z-2 i}{z+2 i}\right)=0\) 

\( \begin{aligned} & \therefore \frac{x^2+y^2-4}{x^2+(y+2)^2}=0 \end{aligned} \) 

\( \begin{aligned} x^2+y^2=4 \text { (circle) } \end{aligned} \) 

Maximum value of |z – (6 + 8i)| means Maximum distance of point (6, 8) from the circle \(\text{x}^2\) + \(\text{y}^2\) = 4 Max. distance = OP + r = 10 + 2 = 12