Show how the current i_3=7e^(jπ/4) can be multiplied by Z_3 (in exponential form) to give the voltage through the branch. Calculate using exponential values, then convert to polar so show value and phase of the voltage.
\$Z_3=15\angle-60^\circ = 15e^{-j \frac{π}{3}}\$
Using the values I got \$(15e^\frac{-j\pi}{3}) ( 7e^{\frac{j\pi}{4}})= 105e^\frac{\pi}{12}=105 \angle 15^{\circ}\$
The correct answer is \$105e^{-j\frac{\pi}{12}}\$
The only wrong thing about your answer is it is missing a negative sign in the exponent.
But it appears you didn't just forget to write down the negative sign. It seems you arrived at that answer in a completely incorrect way and by bad coincidence that both approaches end up with 12 in the denominator of the exponent in the final result, and similar looking, but not the same results.
It appears you multiplied the denominator (and only the denominator) while partially just copying over the numerator without performing any operation. You do not multiply exponents when multiplying exponentials; You add the entirety of the exponents.
Even if you were supposed to multiply the exponent you end up with \$(-j\frac{\pi}{3})(j\frac{\pi}{4}) = \frac{\pi^2}{12}\$. So your incorrect answer is a strange mix of incorrect and inconsistently applied operations.
The correct way is: \$-j\frac{\pi}{3}+j\frac{\pi}{4} = -j\frac{\pi}{12}\$
