solve-for-x-C-3-2ix-3-ix-2-5-0-

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

solve for x \in C

3^{2 i x} - 3^{i x} 2 + 5 = 0

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

{(3^{i x})}^{2} - 2 (3^{i x}) + 1 + 4 = 0

{(3^{i x} - 1)}^{2} = - 4

3^{i x} - 1 = \pm 2 i

3^{i x} = 1 \pm 2 i

\ln (3^{i x}) = \ln (1 \pm 2 i)

x = \frac{\ln (1 \pm 2 i)}{i \ln 3}

x = - \frac{i \ln (1 \pm 2 i)}{\ln 3}

Commented by Ghisom last updated on 20/Feb/24

thank you

Answered by Frix last updated on 20/Feb/24

With x = a + b i; a, b \in R we get :

Real part :

\frac{1}{9^{b}} ({2 \cos}^{2} (a \ln 3) - 3^{b} 2 \cos (a \ln 3) + 9^{b} 5 - 1) = 0

Imaginary part :

\frac{2}{9^{b}} (\cos (a \ln 3) - 3^{b}) \sin (a \ln 3) = 0

\frac{1}{9^{b}} \neq 0 \Rightarrow

{\begin{cases} (1) \cos^{2} (a \ln 3) - 3^{b} \cos (a \ln 3) + \frac{9^{b} 5 - 1}{2} = 0 \\ (2) (\cos (a \ln 3) - 3^{b}) \sin (a \ln 3) = 0 \end{cases}

(2)

With \sin (a \ln 3) = 0 \Rightarrow b \notin R

\cos (a \ln 3) = 3^{b} \Rightarrow

(1) 9^{b} = \frac{1}{5} \Rightarrow b = - \frac{\ln 5}{2 \ln 3}

\Rightarrow \cos (a \ln 3) = \frac{\sqrt{5}}{5} \Rightarrow a = \frac{2 n π \pm \cos^{- 1} \frac{\sqrt{5}}{5}}{\ln 3}

x = \frac{2 n π \pm \cos^{- 1} \frac{\sqrt{5}}{5}}{\ln 3} - \frac{\ln 5}{2 \ln 3} i; n \in Z

Commented by Ghisom last updated on 20/Feb/24

thank you