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If-z-i-2-and-z-0-5-3i-then-the-maximum-value-of-z-0-iz-is-A-7-B-7-C-5-D-9-

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Question Number 139394 by EnterUsername last updated on 26/Apr/21
If ∣z−i∣≤2 and z_0 =5+3i, then the maximum value of ∣z_0 +iz∣ is (A) 7 (B) (√7) (C) 5 (D) 9
$$\mathrm{If}\:\mid{z}−{i}\mid\leqslant\mathrm{2}\:\mathrm{and}\:{z}_{\mathrm{0}} =\mathrm{5}+\mathrm{3}{i},\:\mathrm{then}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value} \\ $$$$\mathrm{of}\:\mid{z}_{\mathrm{0}} +{iz}\mid\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{7}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\sqrt{\mathrm{7}}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{5}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{9} \\ $$
Answered by mr W last updated on 26/Apr/21
z=x+yi ∣z−i∣=(√(x^2 +(y−1)^2 ))≤2 x^2 +(y−1)^2 ≤4 F=∣z_0 +iz∣=∣5−y+(3+x)i∣ =(√((5−y)^2 +(3+x)^2 )) =(√((x+3)^2 +(y−5)^2 )) F_(min) =(√((−3)^2 +(5−1)^2 ))−2=3 F_(max) =(√((−3)^2 +(5−1)^2 ))+2=7
$${z}={x}+{yi} \\ $$$$\mid{z}−{i}\mid=\sqrt{{x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} }\leqslant\mathrm{2} \\ $$$${x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} \leqslant\mathrm{4} \\ $$$$ \\ $$$${F}=\mid{z}_{\mathrm{0}} +{iz}\mid=\mid\mathrm{5}−{y}+\left(\mathrm{3}+{x}\right){i}\mid \\ $$$$\:\:\:=\sqrt{\left(\mathrm{5}−{y}\right)^{\mathrm{2}} +\left(\mathrm{3}+{x}\right)^{\mathrm{2}} } \\ $$$$\:\:\:=\sqrt{\left({x}+\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{5}\right)^{\mathrm{2}} } \\ $$$${F}_{{min}} =\sqrt{\left(−\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{5}−\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{2}=\mathrm{3} \\ $$$${F}_{{max}} =\sqrt{\left(−\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{5}−\mathrm{1}\right)^{\mathrm{2}} }+\mathrm{2}=\mathrm{7} \\ $$
Commented by EnterUsername last updated on 26/Apr/21
How do you get F_(min ) and F_(max) , Sir ?
$$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{get}\:{F}_{\mathrm{min}\:} \:\mathrm{and}\:{F}_{\mathrm{max}} ,\:\mathrm{Sir}\:? \\ $$
Commented by mr W last updated on 26/Apr/21
F is the distance from point (x,y) to point (−3,5). point (x,y) should be on the circle x^2 +(y−1)^2 ≤2^2 , a circle at center (0,1) and with radius 2. so the minimum distance F is the distance from (−3,5) to (0,1) minus radius 2 and the maximum distance F is the distance from (−3,5) to (0,1) plus radius 2.
$${F}\:\:{is}\:{the}\:{distance}\:{from}\:{point}\:\left({x},{y}\right)\:{to} \\ $$$${point}\:\left(−\mathrm{3},\mathrm{5}\right).\:{point}\:\left({x},{y}\right)\:{should}\:{be} \\ $$$${on}\:{the}\:{circle}\:{x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} \leqslant\mathrm{2}^{\mathrm{2}} ,\:{a}\:{circle} \\ $$$${at}\:{center}\:\left(\mathrm{0},\mathrm{1}\right)\:{and}\:{with}\:{radius}\:\mathrm{2}. \\ $$$${so}\:{the}\:{minimum}\:{distance}\:{F}\:{is}\:{the} \\ $$$${distance}\:{from}\:\left(−\mathrm{3},\mathrm{5}\right)\:{to}\:\left(\mathrm{0},\mathrm{1}\right)\:{minus} \\ $$$${radius}\:\mathrm{2}\:{and}\:{the}\:{maximum}\:{distance} \\ $$$${F}\:{is}\:{the}\:{distance}\:{from}\:\left(−\mathrm{3},\mathrm{5}\right)\:{to}\:\left(\mathrm{0},\mathrm{1}\right)\: \\ $$$${plus}\:{radius}\:\mathrm{2}. \\ $$
Commented by EnterUsername last updated on 26/Apr/21
Understood! Thank you Sir.
$$\mathrm{Understood}!\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}. \\ $$

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