If m be the least value of |z-3+4i|^{2}+|z-5-2i|^{2}, z belongs

Answered question

2022-01-17

If m be the least value of |z3+4i|2+|z52i|2,z belongs to C, attained at z=z0 then what does the ordered pair (|z0|,m) is equal to?

Answer & Explanation

nick1337

nick1337

Expert2022-01-17Added 777 answers

Step 1
Minimize
U=|zu|2+|zv|2
where
u=34iv=5+2i
That’s the sum of the squared distance from z to u and from z to v. That has to be minimized somewhere along the segment from u to v. A little thought yields that must the midpoint,
12+12<(1+1)2z0=u+v2=4i|z0u|2=(43)2+(14)2=10
We chose z0 so |z0v| is the same; let's check:
|z0v|2=(45)2+(12)2=10m=10+10=20(|z0|,m)=(17,20)

star233

star233

Skilled2022-01-17Added 403 answers

Step 1
Write your function as
f(z,z¯)=(z3+4i)(z¯34i)+(z52i)(z¯5+2i)
and observe that the partial derivative with respect to z¯ is
fz=2z8+2i
which vanishes for z=z0=4i. Then z¯=4+i and
m=f(z0,z¯0)=|1+3i|2+|13i|2=20

alenahelenash

alenahelenash

Expert2022-01-24Added 556 answers

Let z=x+iy Putting this in the above expression gives, f(x,y)=(x3)2+(y+4)2+(x5)2+(y2)2 This can be further simplified as, f(x,y)=2[(x4)2+(y+1)2]+20 The value of the expression in the brackets has to be zero. Now we have, z=4i for minimizing f(x,y) Minimum value of f(x,y)=20 |z|=(17)12

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