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

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

A complex number can be dictated as:

$\mathit{z}\mathbf{=}\mathit{x}\mathbf{+}\mathit{i}\mathit{y}$

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

The modulus of a complex number can be calculated as:

$\mathbf{\left|}\mathit{z}\mathbf{\right|}\mathbf{=}\mathbf{\surd }\mathbf{(}{\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{)}$

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

$$\begin{gathered} z=x+yi \\ =\frac{1-it}{2t+i}\times \frac{2t-i}{2t=i} \\ =\frac{2t=i-2t^{2}i-t}{4t^2+2} \end{gathered}$$

The value of x and y is given below:

$$\begin{gathered} x=\frac{t}{4t^2+1} \\ y=-\frac{\left(1+2t^2\right)}{4t^2+1} \\ y=\frac{-1}{2}\left(\frac{2+4t^2}{4t^2+1}\right) \\ =\frac{-1}{2}\left(\frac{1}{4t^2+1}+1\right) \end{gathered}$$

The formula for velocity is given below:

$$\begin{gathered} v\left(t\right)=\left|\frac{dz}{dt}\right| \\ v\left(t\right)=\left|\frac{dx}{dt}+i\frac{dy}{dt}\right| \end{gathered}$$

Find $\frac{dx}{dt}$as:

$$\begin{gathered} \frac{dx}{dt}=\frac{\left(4t^2+1\right)-\left(t\right)\left(8t\right)}{{\left(4t^2+1\right)}^2} \\ =\frac{1-4t^2}{{\left(4t^2+1\right)}^2} \end{gathered}$$

Find $\frac{dy}{dt}$ as:

$$\begin{gathered} \frac{dy}{dt}=\left(-\frac{1}{2}\right)\frac{0-\left(8t\right)}{{\left(4t^2+1\right)}^2} \\ =\frac{4t}{{\left(4t^2+1\right)}^2} \end{gathered}$$

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

$$\begin{gathered} 4\left(t\right)=\sqrt{{\left(\frac{4t}{{\left(4t^2+1\right)}^2}\right)}^2+{\left(\frac{1-4t^2}{{\left(4t^2+1\right)}^2}\right)}^2} \\ =\frac{1}{{\left(4t^2+1\right)}^2}\sqrt{\left(1-8t^2+16t^4\right)+{\left(4t\right)}^2} \\ =\frac{1}{{\left(4t^2+1\right)}^2}\sqrt{{\left(1+4t^2\right)}^2} \\ =\frac{1}{{\left(4t^2+1\right)}^2} \end{gathered}$$

The formula for velocity is given below:

$$\begin{gathered} a\left(t\right)=\left|\frac{d^{2}z}{d{t}^2}\right| \\ =\left|\frac{d^{2}x}{d{t}^2}+i\frac{d^{2}x}{d{t}^2}\right| \end{gathered}$$

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

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

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

localid="1658550400226" $$\begin{gathered} \frac{d^{2}y}{d{t}^3}=\frac{d}{dt}\left[\frac{4t}{{\left(4t^2+1\right)}^2}\right] \\ =\left[\frac{4{\left(4t^2+1\right)}^2-\left(4t\right)\left(2\right)\left(4t^2+1\right)\left(8t\right)}{{\left(4t^2+1\right)}^4}\right] \\ =\left[\frac{16t^2+4-64t^2}{{\left(4t^2+1\right)}^3}\right] \\ =\frac{4-48t^2}{{\left(4t^2+1\right)}^3} \end{gathered}$$

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

localid="1658550747252" $$\begin{gathered} a=\sqrt{{\left(\frac{4-48t^2}{{\left(4t^2+1\right)}^3}\right)}^2+{\left(\frac{32t^3-24t}{{\left(4t^2+1\right)}^3}\right)}^2} \\ =\frac{1}{{\left(4t^2+1\right)}^3}\sqrt{{\left(4-48t^2\right)}^2+{\left(32t^3-24t\right)}^2} \\ =\frac{1}{{\left(4t^2+1\right)}^3}\sqrt{4^2{\left(1-12t^2\right)}^2+4{\left(8t^3-6t\right)}^2} \end{gathered}$$

Solve further as:

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

The value of a and v is given below:

$$\begin{gathered} v=\frac{1}{4t^2+1} \\ a==\frac{4}{{\left(4t^2+1\right)}^{3/2}} \end{gathered}$$