Question Number 208892 by hardmath last updated on 26/Jun/24
$$\mathrm{Find}: \\ $$$$\sqrt{−\mathrm{16}}\:\:\centerdot\:\:\sqrt{−\mathrm{9}}\:\:=\:\:? \\ $$
Commented by Adeyemi889 last updated on 26/Jun/24
$$ \\ $$$$\sqrt{−\mathrm{16}}\:=\:\sqrt{\:\left(\mathrm{16}\right)\left(−\mathrm{1}\right)}\:=\sqrt{−\mathrm{1}}\:×\sqrt{\mathrm{16}\:} \\ $$$${note}\:\sqrt{−\mathrm{1}}\:=\:\boldsymbol{{i}} \\ $$$$\sqrt{−\mathrm{16}}\:=\mathrm{4}\boldsymbol{{i}} \\ $$$$ \\ $$
Answered by mr W last updated on 26/Jun/24
$$=\sqrt{−\mathrm{1}}×\sqrt{\mathrm{16}}×\sqrt{−\mathrm{1}}×\sqrt{\mathrm{9}} \\ $$$$=\left(\sqrt{−\mathrm{1}}\right)^{\mathrm{2}} ×\mathrm{4}×\mathrm{3} \\ $$$$=−\mathrm{12} \\ $$
Answered by Frix last updated on 26/Jun/24
$$\sqrt{−\mathrm{16}}=\sqrt{\mathrm{16e}^{\mathrm{i}\pi} }=\mathrm{4e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} \\ $$$$\sqrt{−\mathrm{9}}=\sqrt{\mathrm{9e}^{\mathrm{i}\pi} }=\mathrm{3e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} \\ $$$$\sqrt{−\mathrm{16}}×\sqrt{−\mathrm{9}}=\mathrm{4e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} ×\mathrm{3e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} =\mathrm{12e}^{\mathrm{i}\pi} =−\mathrm{12} \\ $$
Commented by hardmath last updated on 28/Jun/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professors} \\ $$