Answer
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Work Step by Step
When the isotope $^{64}_{29}Cu$ decays by $\gamma$ emission, the resulting nuclide is the exact same isotope, but in a lower energy state.
$$^{64}_{29}Cu^*\rightarrow\;^{64}_{29}Cu +\gamma$$
When the isotope $^{64}_{29}Cu$ decays by $\beta^-$ emission, the resulting nuclide is an isotope of zinc.
$$^{64}_{29}Cu\rightarrow\;^{64}_{30}Zn +e^-+\overline{\nu}$$
When the isotope $^{64}_{29}Cu$ decays by $\beta^+$ emission, the resulting nuclide is an isotope of nickel.
$$^{64}_{29}Cu\rightarrow\;^{64}_{28}Ni +e^+ +\nu$$
