Write e^(ln7+π i) in the form u+iv

Write $e^{\ln7+\pi i}$ in the form $u+iv$.

Explanation of Method

Split via $e^{a+bi}=e^a\cdot e^{bi}$, then evaluate each factor separately: $e^{\ln7}=7$ by the inverse relationship between $\exp$ and $\ln$, and $e^{\pi i}=-1$ by Euler's identity.

Step 1 · Split the exponent

$$ e^{\ln7+\pi i}=e^{\ln7}\cdot e^{\pi i} $$

Step 2 · Evaluate each factor

$$ e^{\ln7}=7, \qquad e^{\pi i}=\cos\pi+i\sin\pi=-1+0i=-1 $$

Step 3 · Combine

$$ e^{\ln7+\pi i}=7\cdot(-1)=-7 $$

$$ \boxed{u=-7, \quad v=0} $$

Verification

Direct check via the definition $e^{x+iy}=e^x(\cos y+i\sin y)$ with $x=\ln7,\ y=\pi$: $\ e^{\ln7}(\cos\pi+i\sin\pi)=7(-1+0i)=-7$ ✓.

Result

$$ e^{\ln7+\pi i}=-7 $$

A direct instance of Euler's identity $e^{\pi i}=-1$, scaled by the factor $e^{\ln7}=7$ — the result is purely real.