What is the value of arcsinh(5+12i) ?

The value of arcsinh(5+12i) is ln(5+12i+sqrt(-118+120i))

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arcsinh(5+12i)

arcsinh (5 + 12 i)

ln (5 + 12 i + - 118 + 120 i)

Solution steps

arcsinh (5 + 12 i)

Use the Hyperbolic identity:

arcsinh (x) = ln (x + x^{2} + 1)

= ln (5 + 12 i + (5 + 12 i)^{2} + 1)

Simplify

ln (5 + 12 i + (5 + 12 i)^{2} + 1) : ln (5 + 12 i + - 118 + 120 i)

ln (5 + 12 i + (5 + 12 i)^{2} + 1)

Expand

(5 + 12 i)^{2} + 1 : - 118 + 120 i

(5 + 12 i)^{2} + 1

(5 + 12 i)^{2} : - 119 + 120 i

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 5, b = 12 i

= 5^{2} + 2 \cdot 5 \cdot 12 i + (12 i)^{2}

Simplify

5^{2} + 2 \cdot 5 \cdot 12 i + (12 i)^{2} : - 119 + 120 i

5^{2} + 2 \cdot 5 \cdot 12 i + (12 i)^{2}

5^{2} = 25

5^{2}

5^{2} = 25

= 25

2 \cdot 5 \cdot 12 i = 120 i

2 \cdot 5 \cdot 12 i

Multiply the numbers:

2 \cdot 5 \cdot 12 = 120

= 120 i

(12 i)^{2} = - 144

(12 i)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 1 2^{2} i^{2}

1 2^{2} = 144

= 144 i^{2}

i^{2} = - 1

i^{2}

Apply imaginary number rule:

i^{2} = - 1

= - 1

= 144 (- 1)

Remove parentheses:

(- a) = - a

= - 144 \cdot 1

Multiply the numbers:

144 \cdot 1 = 144

= - 144

= 25 + 120 i - 144

Subtract the numbers:

25 - 144 = - 119

= - 119 + 120 i

= - 119 + 120 i

= (- 119 + 120 i) + 1

Simplify

(- 119 + 120 i) + 1 : - 118 + 120 i

(- 119 + 120 i) + 1

Remove parentheses:

(- a) = - a

= - 119 + 120 i + 1

Add/Subtract the numbers:

- 119 + 1 = - 118

= - 118 + 120 i

= - 118 + 120 i

= ln (5 + 12 i + - 118 + 120 i)

= ln (5 + 12 i + - 118 + 120 i)

sin (10 2^{\circ})

- 2 sec (\frac{π}{2})

5 + 4 cos (0)

arctan (0.245)

2 arccos (\frac{3}{2})

What is the value of arcsinh(5+12i) ?
The value of arcsinh(5+12i) is ln(5+12i+sqrt(-118+120i))