What is the value of cosh(1+i) ?

The value of cosh(1+i) is (cos(1)+e^2cos(1))/(2e)+i(-sin(1)+e^2sin(1))/(2e)

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cosh(1+i)

Solution

cosh (1 + i)

Solution

\frac{cos ( 1 ) + e ^{2} cos ( 1 )}{2 e} + i \frac{- sin ( 1 ) + e ^{2} sin ( 1 )}{2 e}

Solution steps

cosh (1 + i)

Use the Hyperbolic identity:

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

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

Simplify

\frac{e ^{1 + i} + e ^{- (1 + i)}}{2} : \frac{cos ( - 1 ) + e ^{2} cos ( 1 )}{2 e} + i \frac{sin ( - 1 ) + e ^{2} sin ( 1 )}{2 e}

\frac{e ^{1 + i} + e ^{- (1 + i)}}{2}

e^{1 + i} + e^{- (1 + i)} = e (cos (1) + i sin (1)) + e^{- 1} (cos (- 1) + i sin (- 1))

e^{1 + i} + e^{- (1 + i)}

e^{1 + i} = e (cos (1) + i sin (1))

e^{1 + i}

Apply imaginary number rule:

e^{a + ib} = e^{a} (cos (b) + i sin (b))

= e^{1} (cos (1) + i sin (1))

Apply rule

a^{1} = a

e^{1} = e

= e (cos (1) + i sin (1))

e^{- (1 + i)} = e^{- 1} (cos (- 1) + i sin (- 1))

e^{- (1 + i)}

Apply imaginary number rule:

e^{a + ib} = e^{a} (cos (b) + i sin (b))

= e^{- 1} (cos (- 1) + i sin (- 1))

= e (cos (1) + i sin (1)) + e^{- 1} (cos (- 1) + i sin (- 1))

= \frac{e ( cos ( 1 ) + i sin ( 1 ) ) + e ^{- 1} ( cos ( - 1 ) + i sin ( - 1 ) )}{2}

e^{- 1} (cos (- 1) + sin (- 1) i) = \frac{cos ( - 1 ) + i sin ( - 1 )}{e}

e^{- 1} (cos (- 1) + sin (- 1) i)

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

= \frac{1}{e} (cos (- 1) + i sin (- 1))

Multiply fractions:

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

= \frac{1 \cdot ( cos ( - 1 ) + sin ( - 1 ) i )}{e}

1 \cdot (cos (- 1) + sin (- 1) i) = cos (- 1) + i sin (- 1)

1 \cdot (cos (- 1) + sin (- 1) i)

Multiply:

1 \cdot (cos (- 1) + sin (- 1) i) = (cos (- 1) + sin (- 1) i)

= (cos (- 1) + i sin (- 1))

Remove parentheses:

(a) = a

= cos (- 1) + sin (- 1) i

= \frac{cos ( - 1 ) + i sin ( - 1 )}{e}

= \frac{e ( cos ( 1 ) + i sin ( 1 ) ) + \frac{c o s ( - 1 ) + i s i n ( - 1 )}{e}}{2}

Join

e (cos (1) + sin (1) i) + \frac{cos ( - 1 ) + sin ( - 1 ) i}{e} : \frac{e ^{2} ( cos ( 1 ) + i sin ( 1 ) ) + cos ( - 1 ) + i sin ( - 1 )}{e}

e (cos (1) + sin (1) i) + \frac{cos ( - 1 ) + sin ( - 1 ) i}{e}

Convert element to fraction:

e (cos (1) + i sin (1)) = \frac{e ( cos ( 1 ) + sin ( 1 ) i ) e}{e}

= \frac{e ( cos ( 1 ) + sin ( 1 ) i ) e}{e} + \frac{cos ( - 1 ) + sin ( - 1 ) i}{e}

Since the denominators are equal, combine the fractions:

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

= \frac{e ( cos ( 1 ) + sin ( 1 ) i ) e + cos ( - 1 ) + sin ( - 1 ) i}{e}

e (cos (1) + sin (1) i) e + cos (- 1) + sin (- 1) i = e^{2} (cos (1) + i sin (1)) + cos (- 1) + i sin (- 1)

e (cos (1) + sin (1) i) e + cos (- 1) + sin (- 1) i

e (cos (1) + sin (1) i) e = e^{2} (cos (1) + i sin (1))

e (cos (1) + sin (1) i) e

Apply exponent rule:

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

ee = e^{1 + 1}

= (cos (1) + sin (1) i) e^{1 + 1}

Add the numbers:

1 + 1 = 2

= (cos (1) + sin (1) i) e^{2}

= e^{2} (cos (1) + i sin (1)) + cos (- 1) + i sin (- 1)

= \frac{e ^{2} ( cos ( 1 ) + i sin ( 1 ) ) + cos ( - 1 ) + i sin ( - 1 )}{e}

= \frac{\frac{e ^{2} ( c o s ( 1 ) + i s i n ( 1 ) ) + c o s ( - 1 ) + i s i n ( - 1 )}{e}}{2}

Apply the fraction rule:

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

= \frac{( cos ( 1 ) + sin ( 1 ) i ) e ^{2} + cos ( - 1 ) + sin ( - 1 ) i}{e 2}

Rewrite

\frac{( cos ( 1 ) + sin ( 1 ) i ) e ^{2} + cos ( - 1 ) + sin ( - 1 ) i}{e 2}

in standard complex form:

\frac{e ^{2} cos ( 1 ) + cos ( - 1 )}{2 e} + \frac{e ^{2} sin ( 1 ) + sin ( - 1 )}{2 e} i

\frac{( cos ( 1 ) + sin ( 1 ) i ) e ^{2} + cos ( - 1 ) + sin ( - 1 ) i}{e 2}

Expand

(cos (1) + sin (1) i) e^{2} + cos (- 1) + sin (- 1) i : e^{2} cos (1) + e^{2} i sin (1) + cos (- 1) + sin (- 1) i

(cos (1) + sin (1) i) e^{2} + cos (- 1) + sin (- 1) i

= e^{2} (cos (1) + i sin (1)) + cos (- 1) + i sin (- 1)

Expand

e^{2} (cos (1) + sin (1) i) : e^{2} cos (1) + e^{2} i sin (1)

e^{2} (cos (1) + sin (1) i)

Apply the distributive law:

a (b + c) = ab + a c

a = e^{2}, b = cos (1), c = sin (1) i

= e^{2} cos (1) + e^{2} sin (1) i

= e^{2} cos (1) + e^{2} i sin (1)

= e^{2} cos (1) + e^{2} i sin (1) + cos (- 1) + sin (- 1) i

= \frac{e ^{2} cos ( 1 ) + e ^{2} i sin ( 1 ) + cos ( - 1 ) + i sin ( - 1 )}{2 e}

Apply the fraction rule:

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

\frac{e ^{2} cos ( 1 ) + e ^{2} i sin ( 1 ) + cos ( - 1 ) + i sin ( - 1 )}{2 e} = \frac{e ^{2} cos ( 1 )}{2 e} + \frac{e ^{2} i sin ( 1 )}{2 e} + \frac{cos ( - 1 )}{2 e} + \frac{i sin ( - 1 )}{2 e}

= \frac{e ^{2} cos ( 1 )}{2 e} + \frac{e ^{2} i sin ( 1 )}{2 e} + \frac{cos ( - 1 )}{2 e} + \frac{i sin ( - 1 )}{2 e}

Group like terms

= \frac{cos ( - 1 )}{2 e} + \frac{i sin ( - 1 )}{2 e} + \frac{e ^{2} cos ( 1 )}{2 e} + \frac{e ^{2} i sin ( 1 )}{2 e}

Cancel

\frac{e ^{2} cos ( 1 )}{2 e} : \frac{e cos ( 1 )}{2}

\frac{e ^{2} cos ( 1 )}{2 e}

Cancel the common factor:

e

= \frac{e cos ( 1 )}{2}

= \frac{cos ( - 1 )}{2 e} + \frac{i sin ( - 1 )}{2 e} + \frac{e cos ( 1 )}{2} + \frac{e ^{2} i sin ( 1 )}{2 e}

Cancel

\frac{e ^{2} i sin ( 1 )}{2 e} : \frac{e i sin ( 1 )}{2}

\frac{e ^{2} i sin ( 1 )}{2 e}

Cancel the common factor:

e

= \frac{e i sin ( 1 )}{2}

= \frac{cos ( - 1 )}{2 e} + \frac{i sin ( - 1 )}{2 e} + \frac{e cos ( 1 )}{2} + \frac{e i sin ( 1 )}{2}

Group like terms

= \frac{e cos ( 1 )}{2} + \frac{cos ( - 1 )}{2 e} + \frac{e i sin ( 1 )}{2} + \frac{i sin ( - 1 )}{2 e}

Group the real part and the imaginary part of the complex number

= (\frac{e cos ( 1 )}{2} + \frac{cos ( - 1 )}{2 e}) + (\frac{e sin ( 1 )}{2} + \frac{sin ( - 1 )}{2 e}) i

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

\frac{e sin ( 1 )}{2} + \frac{sin ( - 1 )}{2 e}

Least Common Multiplier of

2, 2 e : 2 e

2, 2 e

Lowest Common Multiplier (LCM)

Least Common Multiplier of

2, 2 : 2

2, 2

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Multiply each factor the greatest number of times it occurs in either

2

2

= 2

Multiply the numbers:

2 = 2

= 2

Compute an expression comprised of factors that appear either in

2

2 e

= 2 e

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

2 e

For

\frac{e sin ( 1 )}{2} :

multiply the denominator and numerator by

e

\frac{e sin ( 1 )}{2} = \frac{e sin ( 1 ) e}{2 e} = \frac{e ^{2} sin ( 1 )}{2 e}

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

Since the denominators are equal, combine the fractions:

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

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

= (\frac{e cos ( 1 )}{2} + \frac{cos ( - 1 )}{2 e}) + \frac{e ^{2} sin ( 1 ) + sin ( - 1 )}{2 e} i

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

\frac{e cos ( 1 )}{2} + \frac{cos ( - 1 )}{2 e}

Least Common Multiplier of

2, 2 e : 2 e

2, 2 e

Lowest Common Multiplier (LCM)

Least Common Multiplier of

2, 2 : 2

2, 2

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Multiply each factor the greatest number of times it occurs in either

2

2

= 2

Multiply the numbers:

2 = 2

= 2

Compute an expression comprised of factors that appear either in

2

2 e

= 2 e

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

2 e

For

\frac{e cos ( 1 )}{2} :

multiply the denominator and numerator by

e

\frac{e cos ( 1 )}{2} = \frac{e cos ( 1 ) e}{2 e} = \frac{e ^{2} cos ( 1 )}{2 e}

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

Since the denominators are equal, combine the fractions:

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

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

= \frac{e ^{2} cos ( 1 ) + cos ( - 1 )}{2 e} + \frac{e ^{2} sin ( 1 ) + sin ( - 1 )}{2 e} i

= \frac{e ^{2} cos ( 1 ) + cos ( - 1 )}{2 e} + \frac{e ^{2} sin ( 1 ) + sin ( - 1 )}{2 e} i

= \frac{cos ( - 1 ) + e ^{2} cos ( 1 )}{2 e} + i \frac{sin ( - 1 ) + e ^{2} sin ( 1 )}{2 e}

Use the following property:

sin (- x) = - sin (x)

sin (- 1) = - sin (1)

= \frac{cos ( - 1 ) + e ^{2} cos ( 1 )}{2 e} + i \frac{- sin ( 1 ) + e ^{2} sin ( 1 )}{2 e}

Use the following property:

cos (- x) = cos (x)

cos (- 1) = cos (1)

= \frac{cos ( 1 ) + e ^{2} cos ( 1 )}{2 e} + i \frac{- sin ( 1 ) + e ^{2} sin ( 1 )}{2 e}

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Frequently Asked Questions (FAQ)

What is the value of cosh(1+i) ?
The value of cosh(1+i) is (cos(1)+e^2cos(1))/(2e)+i(-sin(1)+e^2sin(1))/(2e)