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Question Number 204512 by Ghisom last updated on 20/Feb/24

$$\mathrm{solve}\mathrm{for}x\in \mathbb{C}$$$$3^{2\mathrm{i}x}-3^{\mathrm{i}x}2+5=0$$

Answered by Rasheed.Sindhi last updated on 20/Feb/24

$${\left(3^{ix}\right)}^2-2\left(3^{ix}\right)+1+4=0$$$${(3^{ix}-1)}^2=-4$$$$3^{ix}-1=\pm 2i$$$$3^{ix}=1\pm 2i$$$$\ln \left(3^{ix}\right)=\ln (1\pm 2i)$$$$x=\frac{\ln (1\pm 2i)}{i\ln 3}$$$$x=-\frac{i\ln (1\pm 2i)}{\ln 3}$$

Commented by Ghisom last updated on 20/Feb/24

thank you

Answered by Frix last updated on 20/Feb/24

$$\mathrm{With}x=a+b\mathrm{i};a,b\in \mathbb{R}\mathrm{we}\mathrm{get}:$$$$\mathrm{Real}\mathrm{part}:$$$$\frac{1}{9^b}({2\cos }^2\left(a\ln 3\right)-3^{b}2\cos \left(a\ln 3\right)+9^{b}5-1)=0$$$$\mathrm{Imaginary}\mathrm{part}:$$$$\frac{2}{9^b}(\cos \left(a\ln 3\right)-3^b)\sin \left(a\ln 3\right)=0$$$$\frac{1}{9^b}\ne 0\Rightarrow$$$$\{\begin{array}{l}\left(1\right){\cos }^2\left(a\ln 3\right)-3^b\cos \left(a\ln 3\right)+\frac{9^{b}5-1}{2}=0\\ \left(2\right)(\cos \left(a\ln 3\right)-3^b)\sin \left(a\ln 3\right)=0\end{array}$$$$\left(2\right)$$$$\mathrm{With}\sin \left(a\ln 3\right)=0\Rightarrow b\notin \mathbb{R}$$$$\cos \left(a\ln 3\right)=3^b\Rightarrow$$$$\left(1\right)9^b=\frac{1}{5}\Rightarrow b=-\frac{\ln 5}{2\ln 3}$$$$\Rightarrow \cos \left(a\ln 3\right)=\frac{\sqrt{5}}{5}\Rightarrow a=\frac{2n\pi \pm {\cos }^{-1}\frac{\sqrt{5}}{5}}{\ln 3}$$$$$$$$x=\frac{2n\pi \pm {\cos }^{-1}\frac{\sqrt{5}}{5}}{\ln 3}-\frac{\ln 5}{2\ln 3}\mathrm{i};n\in \mathbb{Z}$$

Commented by Ghisom last updated on 20/Feb/24

thank you

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