Let x=cosh⁻¹(1+5i), then cosh(x)=1+5i.

cosh(x) is defined as (eˣ+e⁻ˣ)/2=(e²ˣ+1)/(2eˣ).

Therefore, e²ˣ+1=2eˣ(1+5i),

e²ˣ-2eˣ(1+5i)=-1,

Completing the square:

e²ˣ-2eˣ(1+5i)+(1+5i)²=(1+5i)²-1,

(eˣ-(1+5i))²=1+10i-25-1=5(2i-5).

Let (a+ib)²=a²+2abi-b²≡2i-5, then a²-b²=-5 and ab=1, where a and b are real.

Substituting b=1/a:

a²-1/a²=-5, a⁴+5a²-1=0, a²=(√29-5)/2, making b²=1/a²=(√29+5)/2.

Therefore, a=0.4388, b=2.2787 approx.

a+ib=√(2i-5), eˣ-(1+5i)=(0.4388+2.2787i)√5,

eˣ=1+5i+(0.4388+2.2787i)√5=1+0.4388√5+5i+0.2787i√5,

eˣ=1.9813+10.0954i approx.

Let x=u+iθ so eˣ=eᵘe^(iθ)=eᵘ(cosθ+isinθ).

So:

eᵘcosθ=1.9813, eᵘsinθ=10.0954.

From these, tanθ=10.0954/1.9813=5.0954, θ=1.3770 radians approx.

e²ᵘcos²θ+e²ᵘsin²θ=e²ᵘ=1.9813²+10.0954²=105.8422,

eᵘ=10.2880, u=ln(10.2880)=2.3310, x=2.3310+1.3770i.

Therefore cosh⁻¹(1+5i)=2.3310+1.3770i.