Find and if z=(1-it)2t+i .

The value of a and v is given below:v=14t2+1a=14t2+13/2

Q67P - Mathematical Methods in Physical Sciences

Q67P

Question

Find and if $z = \frac{(1 - i t)}{2 t + i}$ .

Step-by-Step Solution

Verified

Answer

The value of a and v is given below:

v = \frac{1}{4 t^{2} + 1} a = \frac{1}{{(4 t^{2} + 1)}^{3 / 2}}

1Step 1: Given Information

The function $z = \frac{1 - i t}{2 t + i}$ .

2Step 2: Definition of the Complex number

A complex number can be dictated as:

$z = x + i y$

Where z is the complex number, x and y are real numbers, and i is known as iota, whose value is $\sqrt{(- 1)}$ .

The modulus of a complex number can be calculated as:

$| z | = \sqrt (x^{2} + y^{2})$

3Step 3: Find the value of x and y

The function $z = \frac{1 - i t}{2 t + i}$ .

$z = x + y i = \frac{1 - i t}{2 t + i} \times \frac{2 t - i}{2 t = i} = \frac{2 t = i - 2 t^{2} i - t}{4 t^{2} + 2}$

The value of x and y is given below:

$x = \frac{t}{4 t^{2} + 1} y = - \frac{(1 + 2 t^{2})}{4 t^{2} + 1} y = \frac{- 1}{2} (\frac{2 + 4 t^{2}}{4 t^{2} + 1}) = \frac{- 1}{2} (\frac{1}{4 t^{2} + 1} + 1)$

4Step 4: Find the value of the velocity

The formula for velocity is given below:

$v (t) = |\frac{d z}{d t}| v (t) = |\frac{d x}{d t} + i \frac{d y}{d t}|$

Find $\frac{d x}{d t}$ as:

$\frac{d x}{d t} = \frac{(4 t^{2} + 1) - (t) (8 t)}{{(4 t^{2} + 1)}^{2}} = \frac{1 - 4 t^{2}}{{(4 t^{2} + 1)}^{2}}$

Find $\frac{d y}{d t}$ as:

$\frac{d y}{d t} = (- \frac{1}{2}) \frac{0 - (8 t)}{{(4 t^{2} + 1)}^{2}} = \frac{4 t}{{(4 t^{2} + 1)}^{2}}$

Substitute the above values in the formula; the velocity is given below:

$4 (t) = \sqrt{{(\frac{4 t}{{(4 t^{2} + 1)}^{2}})}^{2} + {(\frac{1 - 4 t^{2}}{{(4 t^{2} + 1)}^{2}})}^{2}} = \frac{1}{{(4 t^{2} + 1)}^{2}} \sqrt{(1 - 8 t^{2} + 16 t^{4}) + {(4 t)}^{2}} = \frac{1}{{(4 t^{2} + 1)}^{2}} \sqrt{{(1 + 4 t^{2})}^{2}} = \frac{1}{{(4 t^{2} + 1)}^{2}}$

5Step 5: Find the value of the acceleration

The formula for velocity is given below:

$a (t) = |\frac{d^{2} z}{d t^{2}}| = |\frac{d^{2} x}{d t^{2}} + i \frac{d^{2} x}{d t^{2}}|$

Find $\frac{d^{2} x}{d t^{2}}$ as:

$\frac{d^{2} x}{d t^{2}} = \frac{d}{d t} [\frac{1 - 4 t^{2}}{{(4 t^{2} + 1)}^{2}}] = [\frac{(- 8 t) {(4 t^{2} + 1)}^{2} - (1 - 4 t^{2}) (2) {(4 t^{2} + 1)}^{2} (8)}{{(4 t^{2} + 1)}^{4}}] = [\frac{- 32 t^{3} - 8 t - 16 t + 64 t^{3}}{{(4 t^{2} + 1)}^{3}}] = [\frac{32 t^{3} - 24 t}{{(4 t^{2} + 1)}^{3}}]$

Find $\frac{d^{2} y}{d t^{2}}$ as:

$\frac{d^{2} y}{d t^{3}} = \frac{d}{d t} [\frac{4 t}{{(4 t^{2} + 1)}^{2}}] = [\frac{4 {(4 t^{2} + 1)}^{2} - (4 t) (2) (4 t^{2} + 1) (8 t)}{{(4 t^{2} + 1)}^{4}}] = [\frac{16 t^{2} + 4 - 64 t^{2}}{{(4 t^{2} + 1)}^{3}}] = \frac{4 - 48 t^{2}}{{(4 t^{2} + 1)}^{3}}$

Substitute the above values in the formula. The acceleration is given below:

$a = \sqrt{{(\frac{4 - 48 t^{2}}{{(4 t^{2} + 1)}^{3}})}^{2} + {(\frac{32 t^{3} - 24 t}{{(4 t^{2} + 1)}^{3}})}^{2}} = \frac{1}{{(4 t^{2} + 1)}^{3}} \sqrt{{(4 - 48 t^{2})}^{2} + {(32 t^{3} - 24 t)}^{2}} = \frac{1}{{(4 t^{2} + 1)}^{3}} \sqrt{4^{2} {(1 - 12 t^{2})}^{2} + 4 {(8 t^{3} - 6 t)}^{2}}$

Solve further as:

$a = \frac{1}{{(4 t^{2} + 1)}^{3}} \sqrt{(24 t^{2} + 144 t^{4}) + (64 t^{6} - 96 t^{4} + 36^{2})} = \frac{1}{{(4 t^{2} + 1)}^{3}} \sqrt{(64 t^{6} + 48 t^{4} + 12 t^{2} + 1)} = \frac{1}{{(4 t^{2} + 1)}^{3}} \sqrt{{(4 t^{2} + 1)}^{3}} = \frac{1}{{(4 t^{2} + 1)}^{3 / 2}}$

The value of a and v is given below:

$v = \frac{1}{4 t^{2} + 1} a = = \frac{4}{{(4 t^{2} + 1)}^{3 / 2}}$

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