The number of complex numbers z satisfying z^7=z^ 3 and Re(z)>0 is

Taking moduli, $|z|^7=|z|^3$, so $|z|=0$ or $|z|=1$. The case $z=0$ has $\operatorname{Re}(z)=0$ and is rejected. For $|z|=1$ write $z=e^{i\theta}$; then $\overline z=e^{-i\theta}$ and the equation becomes $e^{7i\theta}=e^{-3i\theta}$, i.e. $e^{10i\theta}=1$, giving $\theta=2\pi k/10$ for $k=0,1,\dots,9$ (10 solutions). Without the restriction the total would be $11$, which is a tempting wrong option. The condition $\cos\theta>0$ keeps $\theta=0,\pm36^\circ,\pm72^\circ$, because $\theta=\pm90^\circ$ gives $\operatorname{Re}z=0$ and the remaining angles lie in the left half-plane. This gives exactly $5$ numbers, so the answer is $5$.