Question Number 198197 by necx122 last updated on 13/Oct/23
$${please}\:{helpe} \\ $$$${sinz}\:=\:\mathrm{2}.\:{Find}\:{z} \\ $$
Commented by mokys last updated on 13/Oct/23
Answered by mr W last updated on 13/Oct/23
$${e}^{{iz}} =\mathrm{cos}\:{z}+{i}\:\mathrm{sin}\:{z} \\ $$$${e}^{−{iz}} =\mathrm{cos}\:{z}−{i}\:\mathrm{sin}\:{z} \\ $$$$\Rightarrow{e}^{{iz}} −{e}^{−{iz}} =\mathrm{2}{i}\:\mathrm{sin}\:{z} \\ $$$$\Rightarrow\mathrm{sin}\:{z}=\frac{{e}^{{iz}} −{e}^{−{iz}} }{\mathrm{2}{i}} \\ $$$$\mathrm{sin}\:{z}=\mathrm{2} \\ $$$$\Rightarrow\frac{{e}^{{iz}} −{e}^{−{iz}} }{\mathrm{2}{i}}=\mathrm{2} \\ $$$$\Rightarrow\left({e}^{{iz}} \right)^{\mathrm{2}} −\mathrm{4}{i}\left({e}^{{iz}} \right)−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{e}^{{iz}} =\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right){i}=\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right){e}^{\left(\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}}\right){i}} \\ $$$$\Rightarrow{iz}=\mathrm{ln}\:\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right)+\left(\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}}\right){i} \\ $$$$\Rightarrow{z}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}}\pm{i}\:\mathrm{ln}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$
Commented by necx122 last updated on 14/Oct/23
$${Thank}\:{you}\:{so}\:{much} \\ $$