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Question Number 199899 by sonukgindia last updated on 11/Nov/23

Answered by Rajpurohith last updated on 11/Nov/23

$$cos\left(z\right)+sin\left(z\right)=k$$$$\Rightarrow \left(\frac{e^{iz}+e^{-iz}}{2}\right)+\left(\frac{e^{iz}-e^{-iz}}{2i}\right)=k$$$$\Rightarrow e^{iz}+e^{-iz}+i(e^{-iz}-e^{iz})=2k$$$$\Rightarrow e^{iz}(1-i)+e^{-iz}(1+i)=2k....\left(1\right)$$$$weknow1+i=\frac{2}{1-i}$$$$putt=e^{iz}(1-i)\Rightarrow t^{-1}=e^{-iz}\left(\frac{1+i}{2}\right)$$$$\Rightarrow \left(1\right)reducestot+2t^{-1}=2k$$$$\Rightarrow t^2-2kt+2=0$$$$\Rightarrow {(t-k)}^2-k^2+2=0$$$$\Rightarrow t=k\pm \sqrt{k^2-2}\Rightarrow e^{iz}(1-i)=k\pm \sqrt{k^2-2}$$$$\Rightarrow e^{iz}=\frac{k\pm \sqrt{k^2-2}}{1-i}$$$$\therefore z=-ilog\left(\frac{k\pm \sqrt{k^2-2}}{1-i}\right)$$$$$$

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