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Complex exercise Level 1 asked in the mathematics teacher group


Exercise:

Show that, cos(iln(a+ibaib))=a2b2a2+b2 where a,bR

Solution: Let z1=a+ib&z2=aib and notice that z2 is the conjugate of z1 hence z2=ˉz1

Thus ifz=a+ib=reiθ then ˉz=aib=reiθ so a+ibaib=reiθreiθ=ei(θ+θ)=ei2θ

We have a+ib=r(cosθ+isinθ) so cosθ=ar=aa2+b2 and sinθ=ba2+b2

L.H.S=cos(iln(a+ibaib))=cos(ilnei2θ)=cos(i(i2θ)lne)=cos(2θ)=cos2θ

        =cos2θsin2θ=a2a2+b2b2a2+b2=a2b2a2+b2=R.H.S

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