Let us first convert this fraction into a complex number. To do that, we realize (to make real) the denominator, by multiplying with its complex conjugate
$$\begin{align}
\frac{1+i}{7+24i} &= \frac{1+i}{7+24i} \times \frac{7-24i}{7-24i}\\[1em]
&= \frac{(1+i)(7-24i)}{(7+24i)(7-24i)}\\[1em]
&= \frac{(1+i)(7-24i)}{7^2 - (24i)^2}\\[1em]
&= \frac{31-17i}{49+576}\\[1em]
&= \frac{31}{625}-\frac{17}{625}i
\end{align}$$
Now that we have a complex number, let's find its modulus and argument.
Let $z = x + i y$ be a complex number. Then,
$$\begin{align}
\text{Modulus}(z) = |z| &= \sqrt{x^2+y^2}\\[2em]
\text{Arg}(z) = \theta &= \begin{cases}
\tan^{-1} \left( \frac{y}{x} \right) &\text{ if } a > 0 \\ \\
\tan^{-1} \left( \frac{y}{x} \right) + 180^\circ &\text{ if } a < 0 \\ \\
90^\circ &\text{ if } a = 0 \text{ and } b > 0 \\ \\
270^\circ &\text{ if } a = 0 \text{ and } b < 0
\end{cases}
\end{align}$$
Video:
https://www.youtube.com/watch?v=Ypnr6Ip2tq4
So, $$\begin{align}
\left | \frac{31}{625}-\frac{17}{625}i \right | &= \sqrt{\left ( \frac{31}{625} \right )^2 + \left ( \frac{17}{625} \right )^2}\\[1em]
&= \color{red}{ \frac{\sqrt 2}{25}}\\[1em]
\hline
\text{Arg} \left ( \frac{31}{625}-\frac{17}{625}i \right ) &= \tan^{-1} \left ( \frac{-17/625}{31/625}\right )\\[1em]
&= \tan^{-1} \left ( \frac{-17}{31}\right )\\[1em]
&\approx \color{red}{331^\circ}
\end{align}$$
You can also try this: http://www.mathportal.org/calculators/complex-numbers-calculator/complex-unary-operations-calculator.php
