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Question Number 142829 by liberty last updated on 06/Jun/21
Find the simplest form for T = (√(1+(√(−3)))) +(√(1−(√(−3))))
$$\:{Find}\:{the}\:{simplest}\:{form}\:{for}\: \\ $$$$\:\:{T}\:=\:\sqrt{\mathrm{1}+\sqrt{−\mathrm{3}}}\:+\sqrt{\mathrm{1}−\sqrt{−\mathrm{3}}}\: \\ $$
Answered by EDWIN88 last updated on 06/Jun/21
Let (√a) +(√(−b)) = (√(1+(√(−3)))) with a>0 ,b>0 ⇒a−b+2i(√(ab)) = 1+i(√3) we get { ((a−b=1)),((2(√(ab)) =(√3))) :} squaring { ((a^2 −2ab+b^2 =1...(1))),((4ab=3...(2))) :} addition of (1) and (2) we have a^2 +2ab+b^2 = 4 ⇒a+b = 2 now we solve { ((a−b=1)),((a+b=2)) :} ⇒ { ((a=(3/2))),((b=(1/2))) :} hence (√(1+(√(−3)))) = (√(3/2)) +(√(−(1/2))) in a similar manner we obtain (√(1−(√(−3)))) = (√(3/2))−(√(−(1/2))) Therefore T= (√(3/2))+(√(−(1/2)))+(√(3/2))−(√(−(1/2))) T = 2(√(3/2)) =(√6) □
$$\mathrm{Let}\:\sqrt{\mathrm{a}}\:+\sqrt{−\mathrm{b}}\:=\:\sqrt{\mathrm{1}+\sqrt{−\mathrm{3}}}\:\mathrm{with}\:\mathrm{a}>\mathrm{0}\:,\mathrm{b}>\mathrm{0} \\ $$$$\Rightarrow\mathrm{a}−\mathrm{b}+\mathrm{2}{i}\sqrt{\mathrm{ab}}\:=\:\mathrm{1}+{i}\sqrt{\mathrm{3}}\:\mathrm{we}\:\mathrm{get}\:\begin{cases}{\mathrm{a}−\mathrm{b}=\mathrm{1}}\\{\mathrm{2}\sqrt{\mathrm{ab}}\:=\sqrt{\mathrm{3}}}\end{cases} \\ $$$$\:\mathrm{squaring}\:\begin{cases}{\mathrm{a}^{\mathrm{2}} −\mathrm{2ab}+\mathrm{b}^{\mathrm{2}} =\mathrm{1}…\left(\mathrm{1}\right)}\\{\mathrm{4ab}=\mathrm{3}…\left(\mathrm{2}\right)}\end{cases} \\ $$$$\mathrm{addition}\:\mathrm{of}\:\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right)\:\mathrm{we}\:\mathrm{have}\: \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{2ab}+\mathrm{b}^{\mathrm{2}} \:=\:\mathrm{4}\:\Rightarrow\mathrm{a}+\mathrm{b}\:=\:\mathrm{2} \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{solve}\:\begin{cases}{\mathrm{a}−\mathrm{b}=\mathrm{1}}\\{\mathrm{a}+\mathrm{b}=\mathrm{2}}\end{cases}\:\Rightarrow\begin{cases}{\mathrm{a}=\frac{\mathrm{3}}{\mathrm{2}}}\\{\mathrm{b}=\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$$$\mathrm{hence}\:\sqrt{\mathrm{1}+\sqrt{−\mathrm{3}}}\:=\:\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\:+\sqrt{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\mathrm{in}\:\mathrm{a}\:\mathrm{similar}\:\mathrm{manner}\:\mathrm{we}\:\mathrm{obtain}\: \\ $$$$\sqrt{\mathrm{1}−\sqrt{−\mathrm{3}}}\:=\:\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}−\sqrt{−\frac{\mathrm{1}}{\mathrm{2}}}\: \\ $$$$\mathrm{Therefore}\:\mathrm{T}=\:\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}+\cancel{\sqrt{−\frac{\mathrm{1}}{\mathrm{2}}}}+\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}−\cancel{\sqrt{−\frac{\mathrm{1}}{\mathrm{2}}}}\: \\ $$$$\mathrm{T}\:=\:\mathrm{2}\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\:=\sqrt{\mathrm{6}}\:\Box \\ $$
Commented by liberty last updated on 06/Jun/21
thank you
$${thank}\:{you} \\ $$
Answered by mr W last updated on 06/Jun/21
1+(√(−3))=1+(√3)i=2e^((πi)/3) ⇒(√(1+(√(−3))))=(√2)e^((πi)/6) similarly ⇒(√(1−(√(−3))))=(√2)e^(−((πi)/6)) (√(1+(√(−3))))+(√(1−(√(−3))))=(√2)(e^((πi)/6) +e^(−((πi)/6)) ) =(√2)×2×cos (π/6)=(√2)×2×((√3)/2)=(√6)
$$\mathrm{1}+\sqrt{−\mathrm{3}}=\mathrm{1}+\sqrt{\mathrm{3}}{i}=\mathrm{2}{e}^{\frac{\pi{i}}{\mathrm{3}}} \\ $$$$\Rightarrow\sqrt{\mathrm{1}+\sqrt{−\mathrm{3}}}=\sqrt{\mathrm{2}}{e}^{\frac{\pi{i}}{\mathrm{6}}} \\ $$$${similarly} \\ $$$$\Rightarrow\sqrt{\mathrm{1}−\sqrt{−\mathrm{3}}}=\sqrt{\mathrm{2}}{e}^{−\frac{\pi{i}}{\mathrm{6}}} \\ $$$$\sqrt{\mathrm{1}+\sqrt{−\mathrm{3}}}+\sqrt{\mathrm{1}−\sqrt{−\mathrm{3}}}=\sqrt{\mathrm{2}}\left({e}^{\frac{\pi{i}}{\mathrm{6}}} +{e}^{−\frac{\pi{i}}{\mathrm{6}}} \right) \\ $$$$=\sqrt{\mathrm{2}}×\mathrm{2}×\mathrm{cos}\:\frac{\pi}{\mathrm{6}}=\sqrt{\mathrm{2}}×\mathrm{2}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\sqrt{\mathrm{6}} \\ $$
Commented by liberty last updated on 06/Jun/21
thank you
$${thank}\:{you} \\ $$
Answered by mathmax by abdo last updated on 06/Jun/21
T=(√(1+(√(−3))+(√(1−(√(−3)))))) we have 1+(√(−3))=1+i(√3)=2((1/2)+i((√3)/2))=2e^((iπ)/3) ⇒ (√(1+(√(−3))))=(√2)e^((iπ)/6) also 1−(√(−3))=1−i(√3)=2e^(−((iπ)/3)) ⇒(√(1−(√(−3))))=(√2)e^(−((iπ)/6)) ⇒ T=(√2)(e^((iπ)/6) +e^(−((iπ)/6)) ) =(√2)(2cos((π/6)))=2(√2).((√3)/2) =(√6)
$$\mathrm{T}=\sqrt{\mathrm{1}+\sqrt{−\mathrm{3}}+\sqrt{\mathrm{1}−\sqrt{−\mathrm{3}}}} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{1}+\sqrt{−\mathrm{3}}=\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \:\Rightarrow \\ $$$$\sqrt{\mathrm{1}+\sqrt{−\mathrm{3}}}=\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{6}}} \\ $$$$\mathrm{also}\:\mathrm{1}−\sqrt{−\mathrm{3}}=\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}=\mathrm{2e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \:\Rightarrow\sqrt{\mathrm{1}−\sqrt{−\mathrm{3}}}=\sqrt{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{6}}} \:\Rightarrow \\ $$$$\mathrm{T}=\sqrt{\mathrm{2}}\left(\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{6}}} \:+\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{6}}} \right)\:=\sqrt{\mathrm{2}}\left(\mathrm{2cos}\left(\frac{\pi}{\mathrm{6}}\right)\right)=\mathrm{2}\sqrt{\mathrm{2}}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\sqrt{\mathrm{6}} \\ $$

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