please include the steps and fornulas used if you can because i keep getting the wrong answer, thank you

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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),


Completing the square:



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=1.9813+10.0954i approx.

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


eᵘcosθ=1.9813, eᵘsinθ=10.0954.

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


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

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

ago by Top Rated User (839k points)

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