Exercise:
Show that, cos(iln(a+iba−ib))=a2−b2a2+b2 where a,b∈R
Solution: Let z1=a+ib&z2=a−ib and notice that z2 is the conjugate of z1 hence z2=ˉz1
Thus ifz=a+ib=reiθ then ˉz=a−ib=re−iθ so a+iba−ib=reiθre−iθ=ei(θ+θ)=ei2θ
We have a+ib=r(cosθ+isinθ) so cosθ=ar=a√a2+b2 and sinθ=b√a2+b2
L.H.S=cos(iln(a+iba−ib))=cos(ilnei2θ)=cos(i(i2θ)lne)=cos(−2θ)=cos2θ
=cos2θ−sin2θ=a2a2+b2−b2a2+b2=a2−b2a2+b2=R.H.S
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