Write 2^(2i) in the form u+iv
Write $2^{2i}$ in the form $u+iv$.
Explanation of Method
For $a>0$ and complex $z$, define $a^z=e^{z\ln a}$. Here the exponent $2i\ln2$ is purely imaginary, so the result lands on the unit circle.
Step 1 · Rewrite as an exponential
$$ 2^{2i}=e^{2i\ln2}=e^{i(2\ln2)} $$
Step 2 · Apply Euler's formula
$$ 2^{2i}=\cos(2\ln2)+i\sin(2\ln2) $$
Step 3 · Numerical evaluation
$$ \ln2\approx0.693147, \qquad 2\ln2\approx1.386294\text{ rad} $$
$$ u=\cos(2\ln2)\approx0.183457, \qquad v=\sin(2\ln2)\approx0.983028 $$
Step 4 · Result
$$ \boxed{2^{2i}=\cos(2\ln2)+i\sin(2\ln2)\approx0.1835+0.9830i} $$
Verification
Modulus check: $|2^{2i}|=|e^{i(2\ln2)}|=1$, and indeed $u^2+v^2\approx0.1835^2+0.9830^2\approx0.03365+0.96629\approx1.0000$ ✓. More generally, for any $a>0,\ b\in\mathbb R$: $\ a^{ib}=\cos(b\ln a)+i\sin(b\ln a)$ always has modulus $1$.
Result
A pure phase: raising a positive real to a purely imaginary power always produces a point on the unit circle.