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Question Number 111159 by Rasheed.Sindhi last updated on 02/Sep/20

z is a complex number with Re(z) , Im(z)∈N. Determine z if z.z^− =1000

$${z}\:{is}\:{a}\:{complex}\:{number}\:{with}\: \\ $$$${Re}\left({z}\right)\:,\:{Im}\left({z}\right)\in\mathbb{N}. \\ $$$${Determine}\:{z}\:\:{if} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{z}.\overset{−} {{z}}=\mathrm{1000} \\ $$

Answered by Sarah85 last updated on 02/Sep/20

(Re(z))^2 +(Im(z))^2 =1000 ⇒ z=10+30i∨18+26i∨26+18i∨30+10i

$$\left({Re}\left({z}\right)\right)^{\mathrm{2}} +\left({Im}\left({z}\right)\right)^{\mathrm{2}} =\mathrm{1000} \\ $$$$\Rightarrow \\ $$$${z}=\mathrm{10}+\mathrm{30i}\vee\mathrm{18}+\mathrm{26i}\vee\mathrm{26}+\mathrm{18i}\vee\mathrm{30}+\mathrm{10i} \\ $$

Commented by Rasheed.Sindhi last updated on 02/Sep/20

THαnX miss! Any process?

$$\mathcal{TH}\alpha{n}\mathcal{X}\:{miss}!\:{Any}\:{process}? \\ $$

Commented by Sarah85 last updated on 02/Sep/20

a=(√(1000−b^2 )) and 1≤b≤31

$${a}=\sqrt{\mathrm{1000}−{b}^{\mathrm{2}} }\:{and}\:\mathrm{1}\leqslant{b}\leqslant\mathrm{31} \\ $$

Commented by Rasheed.Sindhi last updated on 03/Sep/20

Other method which makes the search more narrow.

$${Other}\:{method}\:{which}\:{makes} \\ $$$${the}\:{search}\:{more}\:{narrow}. \\ $$

Commented by Rasheed.Sindhi last updated on 03/Sep/20

See Q#110895

$${See}\:{Q}#\mathrm{110895} \\ $$

Answered by Rasheed.Sindhi last updated on 04/Sep/20

z=a+ib z.z^− =1000⇒a^2 +b^2 =1000 1000 is doubly even number, i-e it′s divisible by 4.This implies that a & b are both even : 1000 ∈E⇒a^2 +b^2 ∈E ⇒(a^2 ,b^2 ∈E) ∨ (a^2 ,b^2 ∈O) ⇒(a,b∈E) ∨ (a,b∈O) But (a^2 +b^2 ) is doubly even so a,b∉O: let a=2p+1, b=2q+1 a^2 +b^2 =(2p+1)^2 +(2q+1)^2 =4(p^2 +q^2 +p+q)+2 (singly even) ∴ a,b∈E This narrows the search.Only we have to look for 2,4,6,...,30 now To make the search more narrow: Let a=10t_a +u_a , b=10t_b +u_b a^2 +b^2 =(10t_a +u_a )^2 +(10t_b +u_b )^2 (100t_a ^2 +20t_a u_a +u_a ^2 )+(100t_b ^2 +20t_b u_b +u_b ^2 )=1000 ⇒Unit-digit of u_a ^2 +u_b ^2 =Unit-digit of 1000=0 Only possibilities are: 0^2 +0^2 =0 1^2 +3^2 =10(excluded due to 1,3∈O) , 2^2 +6^2 =40 3^2 +9^2 =90(excluded due to 1,3∈O) 2^2 +4^2 =20 5^2 +5^2 =50(excluded due to 1,3∈O) 6^2 +8^2 =100 possible units:0,2,4,6,8 No help from this second part we have to look for all even numbers upto 30. So finally, z=10+30i∨30+10i∨18+26i∨26+18i

$${z}={a}+{ib} \\ $$$${z}.\overset{−} {{z}}=\mathrm{1000}\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{1000} \\ $$$$\mathrm{1000}\:{is}\:{doubly}\:{even}\:{number}, \\ $$$${i}-{e}\:{it}'{s}\:{divisible}\:{by}\:\mathrm{4}.\mathcal{T}{his}\:{implies} \\ $$$${that}\:{a}\:\&\:{b}\:{are}\:{both}\:{even}\:: \\ $$$$\mathrm{1000}\:\in\mathbb{E}\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \in\mathbb{E} \\ $$$$\Rightarrow\left({a}^{\mathrm{2}} ,{b}^{\mathrm{2}} \in\mathbb{E}\right)\:\vee\:\left({a}^{\mathrm{2}} ,{b}^{\mathrm{2}} \in\mathbb{O}\right) \\ $$$$\Rightarrow\left({a},{b}\in\mathbb{E}\right)\:\vee\:\left({a},{b}\in\mathbb{O}\right) \\ $$$${But}\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\:{is}\:{doubly}\:{even} \\ $$$${so}\:{a},{b}\notin\mathbb{O}:\:{let}\:{a}=\mathrm{2}{p}+\mathrm{1},\:{b}=\mathrm{2}{q}+\mathrm{1} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}{q}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\mathrm{4}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{p}+{q}\right)+\mathrm{2}\:\left({singly}\:{even}\right) \\ $$$$\therefore\:{a},{b}\in\mathbb{E} \\ $$$$\:{This}\:\boldsymbol{{narrows}}\:\boldsymbol{{the}}\:\boldsymbol{{search}}.{Only}\: \\ $$$${we}\:{have}\:{to}\:{look}\:{for}\:\mathrm{2},\mathrm{4},\mathrm{6},...,\mathrm{30}\:{now} \\ $$$$\mathcal{T}{o}\:{make}\:{the}\:{search}\:{more}\:{narrow}: \\ $$$${Let}\:{a}=\mathrm{10}{t}_{{a}} +{u}_{{a}} ,\:{b}=\mathrm{10}{t}_{{b}} +{u}_{{b}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left(\mathrm{10}{t}_{{a}} +{u}_{{a}} \right)^{\mathrm{2}} +\left(\mathrm{10}{t}_{{b}} +{u}_{{b}} \right)^{\mathrm{2}} \\ $$$$\left(\mathrm{100}{t}_{{a}} ^{\mathrm{2}} +\mathrm{20}{t}_{{a}} {u}_{{a}} +{u}_{{a}} ^{\mathrm{2}} \right)+\left(\mathrm{100}{t}_{{b}} ^{\mathrm{2}} +\mathrm{20}{t}_{{b}} {u}_{{b}} +{u}_{{b}} ^{\mathrm{2}} \right)=\mathrm{1000} \\ $$$$\Rightarrow{Unit}-{digit}\:{of}\:{u}_{{a}} ^{\mathrm{2}} +{u}_{{b}} ^{\mathrm{2}} \: \\ $$$$\:\:\:\:={Unit}-{digit}\:{of}\:\mathrm{1000}=\mathrm{0} \\ $$$${Only}\:{possibilities}\:{are}:\: \\ $$$$\:\mathrm{0}^{\mathrm{2}} +\mathrm{0}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{1}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} =\mathrm{10}\left({excluded}\:{due}\:{to}\:\mathrm{1},\mathrm{3}\in\mathbb{O}\right)\:, \\ $$$$\mathrm{2}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} =\mathrm{40} \\ $$$$\mathrm{3}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} =\mathrm{90}\left({excluded}\:{due}\:{to}\:\mathrm{1},\mathrm{3}\in\mathbb{O}\right) \\ $$$$\mathrm{2}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} =\mathrm{20} \\ $$$$\mathrm{5}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} =\mathrm{50}\left({excluded}\:{due}\:{to}\:\mathrm{1},\mathrm{3}\in\mathbb{O}\right) \\ $$$$\mathrm{6}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} =\mathrm{100} \\ $$$${possible}\:{units}:\mathrm{0},\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{8} \\ $$$${No}\:{help}\:{from}\:{this}\:{second}\:{part} \\ $$$${we}\:{have}\:{to}\:{look}\:{for}\:{all}\:{even} \\ $$$${numbers}\:{upto}\:\mathrm{30}. \\ $$$${So}\:{finally}, \\ $$$$\:{z}=\mathrm{10}+\mathrm{30}{i}\vee\mathrm{30}+\mathrm{10}{i}\vee\mathrm{18}+\mathrm{26}{i}\vee\mathrm{26}+\mathrm{18}{i} \\ $$

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