prove that ln(1+e^ix) = i(x/2) + ln[2cos(x/2)]?

Question

PLEASE SHOW WORK

Answer 1

Consider e raised to the RHS
e^{i(x/2) + ln[2cos(x/2)]}
= e^ix/2 * 2cos(x/2)
= 2cos(x/2) * [cos(x/2) + i*sin(x/2)]
= 2cos^2 (x/2) + i*sin(x)
= (cosx + 1) + i*sinx
= 1 + cosx+isinx
= 1 + e^ix
= e raised to LHS

Answer 2

Prove the identity ln[1 + e^(ix)] = i(x/2) + ln[2cos(x/2)].

Remember the trig identities.

e^(ix) = cos x + i sin x
cos x = 2cos²(x/2) - 1
sin x = 2sin(x/2) cos(x/2)
____________

Let's start with the left hand side.

Left Hand Side = ln[1 + e^(ix)] = ln[1 + cos x + i sin x]

= ln{1 + (2cos²(x/2) - 1) + i[2sin(x/2) cos(x/2)]}

= ln{2cos²(x/2) + 2i sin(x/2) cos(x/2)}

= ln{2cos(x/2)[cos(x/2) + i sin(x/2)]}

= ln{2cos(x/2)[e^(ix/2)]} = ln[2cos(x/2)] + ln[e^(ix/2)]

= ln[e^(ix/2)] + ln[2cos(x/2)]

= i(x/2) + ln[2cos(x/2)] = Right Hand Side

Answer 3

we have the identity
e^ix= cos x + i sin x
so ln(1 + e^ix)
= ln(1+cos x + i sin x)
= ln[2cos^2 (x/2) +i 2sin (x/2) cos (x/2) ----------------[using the identities cos 2x=2cos^2 x -1 ,and sin 2x=2sin x cos x]
=ln[2cos (x/2) {cos (x/2) + i sin (x/2)}
=ln[2cos (x/2)] + ln[e^(i x/2)]
=ln[2cos (x/2) + i x/2 ---------------[since ln (e^x)=x]
= i(x/2) + ln[2cos (x/2)]

Answer 4

ln(1+e^ix) - ln[2cos(x/2)] = ln[(1+e^ix) / (2cos(x/2)) ]
= ln[(1+e^ix) / (e^i(x/2)+e^-i(x/2)) ]
= ln[e^i(x/2)(e^i(x/2)+e^i(-x/2)) / (e^i(x/2)+e^i(-x/2)) ]
= ln[e^i(x/2)]
= i(x/2)

Answer 5

1=e^(ix/2)e^(-ix/2);
hence
1+e^(ix)
=e^(ix/2)(e^(-ix/2)+e^(ix/2))
=e^(ix/2)(2cos(x/2));

hence take logarithm on both sides, we have desired results. The last step because e^(ix)=cos(x)+isin(x).

Answer 6

oh man.....sorry i cant help