CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If z is any complex number satisfying |z32i|2, then the minimum value of |2z6+5i| is

Open in App
Solution

Let z=x+iy
|z32i|2....(Given equation)
|x+iy32i|2
|x3+i(y2)|2
So its a circle with radius 2 where 1x5 and 0y4
Now consider |2z6+5i|k
|2(x3)+i(2y+5)|k
4(x3)2+(2y+5)2k.....(1)
For minimum value of k either (x3)2=0 or (2y+5)2=0 or both of them are 0
x3=0x=3. This is value is possible since 1x5.
2y+5=0y=5/2. This value is not possible since 0y4. So we need to consider minimum value of y that is 0.
Put x=3 and y=0 in equation (1), we get
0+25=k
k=5

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon