1 nd all complex values z satisfying the given equation cosh

1.     nd all complex values z satisfying the given equation

                       coshz = i     

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Solution

given coshz = i

cosh(z)= cosh(x+iy)= cos(y)cosh(x)+isin(y)sinh(x) = i

cosh(x) = [e^x+e^-x]/2

sinh(x) =[e^x-e^-x]/2

We must solve: cos(y)cosh(x)+isin(y)sinh(x) = 0+i(1)

Then: we must solve:

cos(y)[e^x+e^-x]/2 = 0 (1)

sin(y)[e^x-e^-x]/2 = 1 (2)

From (1)

cos(y) = 0 or [e^x+e^-x] =0

If [e^x+e^-x] = 0 then [e^(2x)+1]/(e^x) = 0 (no solution)


If cos(y) =0 (x = pi/2 +k(pi))

then sin(y) = 1 or sin(y)=-1

Using (2)

[e^x-e^-x]/2 = 1

or

-[e^x-e^-x]/2 = 1

Then:

If [e^x-e^-x]/2 = 1

(e^2x-1)/e^x = 2

e^2x-2e^x-1=0

(e^x)^2-2(e^x)-1=0

taking u=e^x (x = ln(u))

u^2 -2u-1=0

u = [2±8]/2

u = 1±2

u = 1+2

or

u = 1-2 (not positive)

x = ln(1+2)

Then one possibilty for z=x+iy (when sin(y)=1) is:

y = pi/2 + 2k(pi) (k integer)
x = ln(1+2)


If -[e^x-e^-x]/2 = 1 (sin(y)=-1)

[e^(2x)-1]/(e^x) = -2

e^(2x)+2e^x-1 = 0

(e^x)^2+2(e^x)-1=0

taking u=e^x (x = ln(u))

u^2 +2u-1=0

u = [-2±8]/2

u = -1±2

u = -1+2

or

u = -1-2 (not positive)

x = ln(-1+2)

Then one possibilty for z=x+iy (when sin(y)=-1) is:

y = 3pi/2 + 2k(pi) (k integer)
x = ln(-1+2)