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Question Number 95897 by john santu last updated on 28/May/20

If x∈C . find solution of 3+i(√2) = e^(ix)

$$\mathrm{If}\:{x}\in\mathbb{C}\:.\:\mathrm{find}\:\mathrm{solution}\:\mathrm{of}\: \\ $$$$\mathrm{3}+{i}\sqrt{\mathrm{2}}\:=\:{e}^{{ix}} \: \\ $$

Answered by bobhans last updated on 28/May/20

3+i(√2) = e^(ix) z = 3+i(√2) ⇒∣z∣ = (√(9+2)) = (√(11)) arg(z) = tan^(−1) (((√2)/3)) = 0.44 (√(11)) e^(0.44i) = e^(ix) ln((√(11))) + 0.44i = ix 1.199 +0.44i = ix x = ((1.199+0.44i)/i)= 0.44−1.199i

$$\mathrm{3}+{i}\sqrt{\mathrm{2}}\:=\:{e}^{{ix}} \: \\ $$$${z}\:=\:\mathrm{3}+{i}\sqrt{\mathrm{2}}\:\Rightarrow\mid{z}\mid\:=\:\sqrt{\mathrm{9}+\mathrm{2}}\:=\:\sqrt{\mathrm{11}} \\ $$$$\mathrm{arg}\left({z}\right)\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\right)\:=\:\mathrm{0}.\mathrm{44} \\ $$$$\sqrt{\mathrm{11}}\:\mathrm{e}^{\mathrm{0}.\mathrm{44}{i}} \:=\:{e}^{{ix}} \: \\ $$$$\mathrm{ln}\left(\sqrt{\mathrm{11}}\right)\:+\:\mathrm{0}.\mathrm{44}{i}\:=\:{ix}\: \\ $$$$\mathrm{1}.\mathrm{199}\:+\mathrm{0}.\mathrm{44}{i}\:=\:{ix}\: \\ $$$${x}\:=\:\frac{\mathrm{1}.\mathrm{199}+\mathrm{0}.\mathrm{44}{i}}{{i}}=\:\mathrm{0}.\mathrm{44}−\mathrm{1}.\mathrm{199}{i}\: \\ $$

Answered by abdomathmax last updated on 28/May/20

we have ∣3+i(√2)∣ =(√(11)) ⇒3+i(√2)=(√(11))e^(iarctan(((√2)/3))) (e)⇒e^(ix) =(√(11))e^(iarctan(((√2)/3))) ⇒ix =ln((√(11)))+iarctan(((√2)/3)) ⇒x =−iln((√(11)))+arctan(((√2)/3))

$$\mathrm{we}\:\mathrm{have}\:\mid\mathrm{3}+\mathrm{i}\sqrt{\mathrm{2}}\mid\:=\sqrt{\mathrm{11}}\:\Rightarrow\mathrm{3}+\mathrm{i}\sqrt{\mathrm{2}}=\sqrt{\mathrm{11}}\mathrm{e}^{\mathrm{iarctan}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\right)} \\ $$$$\left(\mathrm{e}\right)\Rightarrow\mathrm{e}^{\mathrm{ix}} \:=\sqrt{\mathrm{11}}\mathrm{e}^{\mathrm{iarctan}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\right)} \:\Rightarrow\mathrm{ix}\:=\mathrm{ln}\left(\sqrt{\mathrm{11}}\right)+\mathrm{iarctan}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\right) \\ $$$$\Rightarrow\mathrm{x}\:=−\mathrm{iln}\left(\sqrt{\mathrm{11}}\right)+\mathrm{arctan}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\right) \\ $$

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