What is the general solution for tanh(z)+2=0 ?

The general solution for tanh(z)+2=0 is No Solution for z\in\mathbb{R}

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Popular >

tanh(z)+2=0

Solution

tanh (z) + 2 = 0

Solution

No Solution for z \in R

Solution steps

tanh (z) + 2 = 0

Rewrite using trig identities

tanh (z) + 2 = 0

Use the Hyperbolic identity:

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

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

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

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

No Solution for

z \in R

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

Multiply both sides by

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

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

Simplify

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

Apply exponent rules

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

Apply exponent rule:

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

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

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

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

Rewrite the equation with

e^{z} = u

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

Solve

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

No Solution for

u \in R

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

Refine

u - \frac{1}{u} + 2 (u + \frac{1}{u}) = 0

Multiply both sides by

u

u - \frac{1}{u} + 2 (u + \frac{1}{u}) = 0

Multiply both sides by

u

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

Simplify

uu - \frac{1}{u} u + 2 (u + \frac{1}{u}) 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 + \frac{1}{u}) u = 0

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

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

Expand

u^{2} - 1 + 2 (u + \frac{1}{u}) u : 3 u^{2} + 1

u^{2} - 1 + 2 (u + \frac{1}{u}) u

= u^{2} - 1 + 2 u (u + \frac{1}{u})

Expand

2 u (u + \frac{1}{u}) : 2 u^{2} + 2

2 u (u + \frac{1}{u})

Apply the distributive law:

a (b + c) = ab + a c

a = 2 u, b = u, c = \frac{1}{u}

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

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

Simplify

2 uu + 2 \cdot \frac{1}{u} u : 2 u^{2} + 2

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

2 uu = 2 u^{2}

2 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= 2 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 2 u^{2}

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

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

Multiply fractions:

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

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

Cancel the common factor:

u

= 1 \cdot 2

Multiply the numbers:

1 \cdot 2 = 2

= 2

= 2 u^{2} + 2

= 2 u^{2} + 2

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

Simplify

u^{2} - 1 + 2 u^{2} + 2 : 3 u^{2} + 1

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

Group like terms

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

Add similar elements:

u^{2} + 2 u^{2} = 3 u^{2}

= 3 u^{2} - 1 + 2

Add/Subtract the numbers:

- 1 + 2 = 1

= 3 u^{2} + 1

= 3 u^{2} + 1

3 u^{2} + 1 = 0

Solve

3 u^{2} + 1 = 0 :

No Solution for

u \in R

3 u^{2} + 1 = 0

Move

1

to the right side

3 u^{2} + 1 = 0

Subtract

1

from both sides

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

Simplify

3 u^{2} = - 1

3 u^{2} = - 1

Divide both sides by

3

3 u^{2} = - 1

Divide both sides by

3

\frac{3 u ^{2}}{3} = \frac{- 1}{3}

Simplify

u^{2} = - \frac{1}{3}

u^{2} = - \frac{1}{3}

x^{2}

cannot be negative for

x \in R

No Solution for u \in R

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 tanh(z)+2=0 ?
The general solution for tanh(z)+2=0 is No Solution for z\in\mathbb{R}