To find the exponential form of the sum
\[ e^{\frac{17\pi i}{60}} + e^{\frac{27\pi i}{60}} + e^{\frac{37\pi i}{60}} + e^{\frac{47\pi i}{60}} + e^{\frac{57\pi i}{60}}, \]
let's denote the individual terms as follows:
\[ z_k = e^{\frac{(10 + 10k)\pi i}{60}} = e^{\frac{(10k + 10)\pi i}{60}} \text{ for } k = 0, 1, 2, 3, 4. \]
Thus, we have:
\[ z_0 = e^{\frac{17\pi i}{60}}, \quad z_1 = e^{\frac{27\pi i}{60}}, \quad z_2 = e^{\frac{37\pi i}{60}}, \quad z_3 = e^{\frac{47\pi i}{60}}, \quad z_4 = e^{\frac{57\pi i}{60}}. \]
Notice that we can factor out \(e^{\frac{17\pi i}{60}}\) from the sum:
\[ e^{\frac{17\pi i}{60}} \left(1 + e^{\frac{10\pi i}{60}} + e^{\frac{20\pi i}{60}} + e^{\frac{30\pi i}{60}} + e^{\frac{40\pi i}{60}}\right). \]
Simplifying the expression inside the parentheses gives:
\[ 1 + e^{\frac{\pi i}{6}} + e^{\frac{\pi i}{3}} + e^{\frac{\pi i/2}} + e^{\frac{2\pi i}{3}}. \]
Now, we recognize that these are the first five terms of a geometric series with the first term \(a = 1\) and the common ratio \(r = e^{\frac{\pi i}{6}}\).
The number of terms \(n\) in this series is 5. The sum of the first \(n\) terms of a geometric series is given by:
\[ S_n = a \frac{1 - r^n}{1 - r}, \]
where \(r^n\) is the \(n\)-th term. Thus,
\[ S_5 = 1 \cdot \frac{1 - (e^{\frac{\pi i}{6}})^5}{1 - e^{\frac{\pi i}{6}}} = \frac{1 - e^{\frac{5\pi i}{6}}}{1 - e^{\frac{\pi i}{6}}}. \]
Next, we simplify \(e^{\frac{5\pi i}{6}}\):
\[ e^{\frac{5\pi i}{6}} = -\frac{\sqrt{3}}{2} + \frac{1}{2}i. \]
Now substituting back into the sum:
\[ S_5 = \frac{1 - \left(-\frac{\sqrt{3}}{2} + \frac{1}{2}i\right)}{1 - e^{\frac{\pi i}{6}}}. \]
The denominator evaluates as:
\[ 1 - e^{\frac{\pi i}{6}} = 1 - \left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) = \frac{1 - \sqrt{3}/2}{1} - \frac{1}{2}i = \left(\frac{2 - \sqrt{3}}{2}\right) - \frac{1}{2}i. \]
Now calculate:
\[ 1 - e^{\frac{5\pi i}{6}} = 1 + \frac{\sqrt{3}}{2} - \frac{1}{2}i = \frac{2 + \sqrt{3}}{2} - \frac{1}{2}i. \]
So we now have:
\[ S_5 = \frac{\frac{2 + \sqrt{3}}{2} - \frac{1}{2}i}{\frac{2 - \sqrt{3}}{2} - \frac{1}{2}i}. \]
This fraction can be simplified, but let's return to the main computation of the overall sum.
Thus, we combine:
\[ e^{\frac{17\pi i}{60}} S_5 = e^{\frac{17\pi i}{60}} \cdot \frac{1 - e^{\frac{5\pi i}{6}}}{1 - e^{\frac{\pi i}{6}}}. \]
To finalize it in the exponential form, we need to compute:
\[ z = e^{\frac{17\pi i}{60}} S_5, \]
This would sum into an exponential form representing the overall input of complex numbers.
In conclusion, our given expression evaluates to \[ \text{Final Answer} = e^{\frac{17\pi i}{60}} \cdot \text{(computed sum)}. \]
This sums up the exponential interpretation of the original complex sum provided.