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if |z-i| less than equal to 2. (z1)=5+3i, then the maximum value of |iz+z1| is (z and z1 are complex no.).

Riddhish Bhalodia
5 years ago
Solution in the figure
jagdish singh singh
173 Points
5 years ago
$\hspace{-18}Given |z-i|\leq 2 and z' = 5+3i\;, Then we calculate \max|iz+z'|\\\\\ So |iz+z'| = |i(z-i)+4+3i|\leq |i(z-i)|+|4+3i|\leq |i||z-i|+5\\\\So we get |iz+z'|\leq |z-i|+5=2+5=7\\\\So we get \max|iz+z'| = 7$
SREEKANTH
85 Points
5 years ago
if you take  z=x+iy
|z-i|=sqrt(x^2+(y-1)^2)
x^2+(y-1)^2
x^2+y^2-2y
z1=5+i3
the maximum value of |iz+z1|=|ix-y+5+i3|
=|(5-y)+i(x+3)|
=sqrt(5-y)^2+(x+3)^2)
=sqrt(x^2+y^2+6x+34-10y)
=sqrt(1.99+34+6x-8y)
=sqrt(1.99+34+6(0.9)-8(0.9)
=sqrt(34+7.3-7.2)
s0  maximum value sqrt(34.111)