What is the general solution for cosh(z)+1=0 ?

The general solution for cosh(z)+1=0 is No Solution for z\in\mathbb{R}

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cosh(z)+1=0

Solution

cosh (z) + 1 = 0

Solution

No Solution for z \in R

Solution steps

cosh (z) + 1 = 0

Rewrite using trig identities

cosh (z) + 1 = 0

Use the Hyperbolic identity:

cosh (x) = \frac{e ^{x} + e ^{- x}}{2}

\frac{e ^{z} + e ^{- z}}{2} + 1 = 0

\frac{e ^{z} + e ^{- z}}{2} + 1 = 0

\frac{e ^{z} + e ^{- z}}{2} + 1 = 0 :

No Solution for

z \in R

\frac{e ^{z} + e ^{- z}}{2} + 1 = 0

Multiply both sides by

2

\frac{e ^{z} + e ^{- z}}{2} \cdot 2 + 1 \cdot 2 = 0 \cdot 2

Simplify

e^{z} + e^{- z} + 2 = 0

Apply exponent rules

e^{z} + e^{- z} + 2 = 0

Apply exponent rule:

a^{b c} = (a^{b})^{c}

e^{- z} = (e^{z})^{- 1}

e^{z} + (e^{z})^{- 1} + 2 = 0

e^{z} + (e^{z})^{- 1} + 2 = 0

Rewrite the equation with

e^{z} = u

u + (u)^{- 1} + 2 = 0

Solve

u + u^{- 1} + 2 = 0 : u = - 1

u + u^{- 1} + 2 = 0

Refine

u + \frac{1}{u} + 2 = 0

Multiply both sides by

u

u + \frac{1}{u} + 2 = 0

Multiply both sides by

u

uu + \frac{1}{u} u + 2 u = 0 \cdot u

Simplify

uu + \frac{1}{u} u + 2 u = 0 \cdot u

Simplify

uu : u^{2}

uu

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= u^{1 + 1}

Add the numbers:

1 + 1 = 2

= u^{2}

Simplify

\frac{1}{u} u : 1

\frac{1}{u} u

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot u}{u}

Cancel the common factor:

u

= 1

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

u^{2} + 1 + 2 u = 0

u^{2} + 1 + 2 u = 0

u^{2} + 1 + 2 u = 0

Solve

u^{2} + 1 + 2 u = 0 : u = - 1

u^{2} + 1 + 2 u = 0

Write in the standard form

a x^{2} + b x + c = 0

u^{2} + 2 u + 1 = 0

Solve with the quadratic formula

u^{2} + 2 u + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = 2, c = 1

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot 1 = 0

2^{2} - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

Subtract the numbers:

4 - 4 = 0

= 0

u_{1, 2} = \frac{- 2 \pm 0}{2 \cdot 1}

u = \frac{- 2}{2 \cdot 1}

\frac{- 2}{2 \cdot 1} = - 1

\frac{- 2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

u = - 1

The solution to the quadratic equation is:

u = - 1

u = - 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

u + u^{- 1} + 2

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = - 1

u = - 1

Substitute back

u = e^{z},

solve for

z

Solve

e^{z} = - 1 :

No Solution for

z \in R

e^{z} = - 1

a^{f (z)}

cannot be zero or negative for

z \in R

No Solution for z \in R

No Solution for z \in R

No Solution for z \in R

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Frequently Asked Questions (FAQ)

What is the general solution for cosh(z)+1=0 ?
The general solution for cosh(z)+1=0 is No Solution for z\in\mathbb{R}