Question Number 88263 by M±th+et£s last updated on 09/Apr/20
$${prove}\:{that}\: \\ $$$$\mid\frac{{e}^{{z}} −{e}^{−{z}} }{\mathrm{2}}\mid^{\mathrm{2}} +{cos}^{\mathrm{2}} {y}={sinh}^{\mathrm{2}} {x}\:\:\:\:\:{when}\:{z}={x}+{iy} \\ $$$$ \\ $$
Commented by M±th+et£s last updated on 09/Apr/20
$${please}\:{help} \\ $$
Answered by TANMAY PANACEA. last updated on 09/Apr/20
![e^(x+iy) =e^x .e^(iy) =(coshx+sinhx)(cosy+isiny) =coshxcosy+icoshxsiny+sinhxcosy+isinhxsiny =cosy(coshx+sinhx)+isiny(coshx+sinhx) e^(−(x+iy)) =e^(−x) .e^(−iy) =(coshx−sinhx)(cosy−isiny) =coshxcosy−icoshxsiny−sinhxcosy+isinhxsiny =cosy(coshx−sinhx)−isiny(coshx−sinhx) now ((e^z −e^(−z) )/2)=(1/2)[cosy(2sinhx)+isiny(2coshx)] ∣((e^z −e^(−z) )/2)∣^2 =(cosysinhx)^2 +(sinycoshx)^2 cos^2 y(sinhx)^2 +sin^2 y(coshx)^2 +cos^2 y =cos^2 y[1+(sinhx)^2 ]+sin^2 y(coshx)^2 =cos^2 y[cos^2 hx−sin^2 hx+sin^2 hx]+sin^2 ycos^2 hx =(cos^2 y+sin^2 y)cos^2 hx =cos^2 hx pls check](https://www.tinkutara.com/question/Q88282.png)
$${e}^{{x}+{iy}} ={e}^{{x}} .{e}^{{iy}} =\left({coshx}+{sinhx}\right)\left({cosy}+{isiny}\right) \\ $$$$={coshxcosy}+{icoshxsiny}+{sinhxcosy}+{isinhxsiny} \\ $$$$={cosy}\left({coshx}+{sinhx}\right)+{isiny}\left({coshx}+{sinhx}\right) \\ $$$$ \\ $$$${e}^{−\left({x}+{iy}\right)} ={e}^{−{x}} .{e}^{−{iy}} =\left({coshx}−{sinhx}\right)\left({cosy}−{isiny}\right) \\ $$$$={coshxcosy}−{icoshxsiny}−{sinhxcosy}+{isinhxsiny} \\ $$$$={cosy}\left({coshx}−{sinhx}\right)−{isiny}\left({coshx}−{sinhx}\right) \\ $$$${now} \\ $$$$\frac{{e}^{{z}} −{e}^{−{z}} }{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\left[{cosy}\left(\mathrm{2}{sinhx}\right)+{isiny}\left(\mathrm{2}{coshx}\right)\right] \\ $$$$ \\ $$$$\mid\frac{{e}^{{z}} −{e}^{−{z}} }{\mathrm{2}}\mid^{\mathrm{2}} =\left({cosysinhx}\right)^{\mathrm{2}} +\left({sinycoshx}\right)^{\mathrm{2}} \\ $$$${cos}^{\mathrm{2}} {y}\left({sinhx}\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} {y}\left({coshx}\right)^{\mathrm{2}} +{cos}^{\mathrm{2}} {y} \\ $$$$={cos}^{\mathrm{2}} {y}\left[\mathrm{1}+\left({sinhx}\right)^{\mathrm{2}} \right]+{sin}^{\mathrm{2}} {y}\left({coshx}\right)^{\mathrm{2}} \\ $$$$={cos}^{\mathrm{2}} {y}\left[{cos}^{\mathrm{2}} {hx}−{sin}^{\mathrm{2}} {hx}+{sin}^{\mathrm{2}} {hx}\right]+{sin}^{\mathrm{2}} {ycos}^{\mathrm{2}} {hx} \\ $$$$=\left({cos}^{\mathrm{2}} {y}+{sin}^{\mathrm{2}} {y}\right){cos}^{\mathrm{2}} {hx} \\ $$$$={cos}^{\mathrm{2}} {hx} \\ $$$$\boldsymbol{{pls}}\:\boldsymbol{{check}} \\ $$
Commented by M±th+et£s last updated on 09/Apr/20
$${yes}\:{sir}\:{its}\:{typo}\:{god}\:{bless}\:{you} \\ $$
Commented by TANMAY PANACEA. last updated on 09/Apr/20
$${most}\:{welcome}\:{sir} \\ $$