Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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.

$$\mathrm{Understood}!\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com