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

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

Answered by bobhans last updated on 28/May/20

$$3+i\sqrt{2}=e^{ix}$$$$z=3+i\sqrt{2}\Rightarrow \mid z\mid =\sqrt{9+2}=\sqrt{11}$$$$\arg \left(z\right)={\tan }^{-1}\left(\frac{\sqrt{2}}{3}\right)=0.44$$$$\sqrt{11}{\mathrm{e}}^{0.44i}=e^{ix}$$$$\ln \left(\sqrt{11}\right)+0.44i=ix$$$$1.199+0.44i=ix$$$$x=\frac{1.199+0.44i}{i}=0.44-1.199i$$

Answered by abdomathmax last updated on 28/May/20

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

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