What is the general solution for coth(z)=1 ?

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

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Popular >

coth(z)=1

Solution

coth (z) = 1

Solution

No Solution for z \in R

Solution steps

coth (z) = 1

Rewrite using trig identities

coth (z) = 1

Use the Hyperbolic identity:

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

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

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

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

No Solution for

z \in R

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

Multiply both sides by

e^{z} - e^{- z}

\frac{e ^{z} + e ^{- z}}{e ^{z} - e ^{- z}} (e^{z} - e^{- z}) = 1 \cdot (e^{z} - e^{- z})

Simplify

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

Apply exponent rules

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

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{z} = u

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

Solve

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

No Solution for

u \in R

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

Refine

u + \frac{1}{u} = u - \frac{1}{u}

Subtract

u

from both sides

u + \frac{1}{u} - u = u - \frac{1}{u} - u

Simplify

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

Multiply both sides by

u

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

Multiply both sides by

u

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

Simplify

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

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

- \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

1 = - 1

1 = - 1

1 = - 1

The sides are not equal

No Solution for u \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 coth(z)=1 ?
The general solution for coth(z)=1 is No Solution for z\in\mathbb{R}