# This complex function has only real values, how can i find? [closed]

$$(1+i\tan\alpha)^{1+i\tan\beta}$$ is a complex function which possess only real values.And one of the real values is $$(\sec\alpha)^{\sec\beta}$$. How to start I am confused. With taking logarithm I tried but I stuck. I need help.

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• Find what? If you set $\beta=0$ your function is $1+i\tan \alpha$ which obviously gives complex values too. – gammatester Dec 6 at 16:44
• Are $\alpha$ and $\beta$ constants? If so what is the variable in the "function". If $\alpha$ is constant and $\beta$ a variable then $\beta = 0$ means that $1 + i\tan \alpha$ is real so $\tan \alpha = 0$ and you have the constant function $f(\beta)=1$. If $\beta$ is constant and $\alpha$ variable you get similar issues. – fleablood Dec 6 at 17:16
• Using $1+i \tan \alpha =\frac{e^{i \alpha }}{\cos \alpha }$ it is possible to write the function in the form $A*B^i$ where $A$ and $B$ are real. Next we will have to see when $B^i$ is real – Lozenges Dec 6 at 17:41

Since the period of $$\tan \alpha$$ is $$\pi$$, without lose of generality we assume that $$-{\pi \over 2}< \alpha<{\pi \over 2}$$. The cases $$\alpha =\pm {\pi \over 2}$$ are special. From which we obtain $$1+i\tan \alpha ={\sqrt{1+\tan^2\alpha}}\cdot e^{i\alpha}={\sec \alpha}\cdot e^{i\alpha}={\sec \alpha}\cdot e^{i(\alpha+2k\pi)}\quad,\quad k\in \Bbb Z$$by substituting in the function we have $$f(\alpha , \beta){=(1+i\tan\alpha)^{1+i\tan\beta}\\=\left( {\sec \alpha}\cdot e^{i(\alpha+2k\pi )}\right)^{1+i\tan \beta}\\=\left( {\sec \alpha}\cdot e^{i(\alpha+2k\pi )}\right)\cdot \sec \alpha ^{i\tan \beta}\cdot e^{-\tan\beta\cdot (\alpha +2k\pi)}\\=\sec \alpha \cdot e^{i(\alpha+2k\pi +\tan \beta \cdot \ln \sec \alpha)}\cdot e^{-\tan\beta\cdot (\alpha +2k\pi)}}$$the only constraint for $$f(\alpha ,\beta)$$ to be real is that $$\alpha+2k\pi +\tan \beta \cdot \ln \sec \alpha=l\pi \quad,\quad l,k\in \Bbb Z$$which means that $${\alpha +\tan \beta \cdot \ln \sec \alpha\over \pi }\in \Bbb Z$$