Question

If \(z = \sum _{n = 0}^{2026}i^n\), where \(i = \sqrt{-1}\), then one of the values of \(\sqrt{z}\) is...

• A. \(e^{i\frac{π}{4}}\)
• B. \(\frac{1}{\sqrt{2}}e^{i\frac{π}{4}}\)
• C. \(e^{i\frac{π}{2}}\)
• D. \(\frac{1}{\sqrt{2}}e^{i\frac{π}{2}}\)

Hint:

Powers of i repeat every four terms and each block of four sums to zero.

Answer 1

The Correct Option is A

Step 1: Group the terms
Use the formula for a geometric series: $z = \dfrac{1 - i^{2027}}{1 - i}$ with 2027 terms and ratio $i$.

Step 2: Reduce the power
$2027 = 4\times506 + 3$, so $i^{2027} = i^3 = -i$. Then $z = \dfrac{1 + i}{1 - i}$.

Step 3: Simplify
Multiply top and bottom by $1 + i$: $z = \dfrac{(1+i)^2}{2} = \dfrac{2i}{2} = i$.

Step 4: Square root in polar form
$i$ has modulus 1 and argument $\pi/2$, so $\sqrt{i}$ has modulus 1 and argument $\pi/4$. Check: $(e^{i\pi/4})^2 = e^{i\pi/2} = i$. The factor $1/\sqrt{2}$ in options B and D would give a modulus of $1/\sqrt{2}$, whose square is $1/2$, not $|z| = 1$.

Final Answer:
One square root of i is e^{i pi/4}. This is option (A). \[ \boxed{\text{(A) }e^{i\pi/4}} \]