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JEE Advanced · Mathematics · 3. Complex Numbers

Let S be the set of all complex numbers z satisfying z-2+i5. If the complex number z0 is such that 1z0-1 is the maximum of the set 1z-1:zS, then the principal argument of 4-z0-z¯0z0-z¯0+2i is

  1. A π2
  2. B π4
  3. C 3π4
  4. D -π2
Verified Solution

Answer & Solution

Correct Answer

(D) -π2

Step-by-step Solution

Detailed explanation


The region represented by z-2+i5 will be region outside and on circle with centre 2,-1 and radius 5.
z-2+i5
Let z=x+iy
x+iy-2+i5
x-2+iy+15
x-22+y+125
x-22+y+125
Now, for 1z0-1 to be maximum, z0-1 must be minimum.
We need to find the Bz0 which is in given region and nearest to point A1,0, hence nearest point from A1,0 will be on the line joining A and C.
Method 1
Let z0=x+iy, then x<2 and y>0 (from diagram)
Consider
\(w=\frac{4-z_0-\bar{z}_0}{z_0-\bar{z}_0+2 i}=\frac{4-(x+i y)-(x-i y)}{(x+i y)-(x-i y)+2 i}=\) \(\frac{4-2 x}{i(y+2)}=\frac{1}{i}\left(\frac{4-x}{y+2}\right)\)
w=-i4-xy+2
w=iK, where K=4-xy+2
As x<2 and y>0K>0
w=-iK, will lie on negative imaginary axis
Argw=-π2
Alternate: As Bz0 lies on line AC,
Equation of AC: y+1=0--11-2x-2
y+1=1-1x-2
x+y=2 ....i
Let z0=x+iy
Consider
w=4-z0-z¯0z0-z¯0+2i
w=4-xiy+2
From equation i
w=4-xi2-x+2
w=1i
w=-i
Argw=-π2
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