Question Number 208892 by hardmath last updated on 26/Jun/24
![Find: (√(−16)) ∙ (√(−9)) = ?](https://www.tinkutara.com/question/Q208892.png)
$$\mathrm{Find}: \\ $$$$\sqrt{−\mathrm{16}}\:\:\centerdot\:\:\sqrt{−\mathrm{9}}\:\:=\:\:? \\ $$
Commented by Adeyemi889 last updated on 26/Jun/24
![(√(−16)) = (√( (16)(−1))) =(√(−1)) ×(√(16 )) note (√(−1)) = i (√(−16)) =4i](https://www.tinkutara.com/question/Q208893.png)
$$ \\ $$$$\sqrt{−\mathrm{16}}\:=\:\sqrt{\:\left(\mathrm{16}\right)\left(−\mathrm{1}\right)}\:=\sqrt{−\mathrm{1}}\:×\sqrt{\mathrm{16}\:} \\ $$$${note}\:\sqrt{−\mathrm{1}}\:=\:\boldsymbol{{i}} \\ $$$$\sqrt{−\mathrm{16}}\:=\mathrm{4}\boldsymbol{{i}} \\ $$$$ \\ $$
Answered by mr W last updated on 26/Jun/24
![=(√(−1))×(√(16))×(√(−1))×(√9) =((√(−1)))^2 ×4×3 =−12](https://www.tinkutara.com/question/Q208894.png)
$$=\sqrt{−\mathrm{1}}×\sqrt{\mathrm{16}}×\sqrt{−\mathrm{1}}×\sqrt{\mathrm{9}} \\ $$$$=\left(\sqrt{−\mathrm{1}}\right)^{\mathrm{2}} ×\mathrm{4}×\mathrm{3} \\ $$$$=−\mathrm{12} \\ $$
Answered by Frix last updated on 26/Jun/24
![(√(−16))=(√(16e^(iπ) ))=4e^(i(π/2)) (√(−9))=(√(9e^(iπ) ))=3e^(i(π/2)) (√(−16))×(√(−9))=4e^(i(π/2)) ×3e^(i(π/2)) =12e^(iπ) =−12](https://www.tinkutara.com/question/Q208901.png)
$$\sqrt{−\mathrm{16}}=\sqrt{\mathrm{16e}^{\mathrm{i}\pi} }=\mathrm{4e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} \\ $$$$\sqrt{−\mathrm{9}}=\sqrt{\mathrm{9e}^{\mathrm{i}\pi} }=\mathrm{3e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} \\ $$$$\sqrt{−\mathrm{16}}×\sqrt{−\mathrm{9}}=\mathrm{4e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} ×\mathrm{3e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} =\mathrm{12e}^{\mathrm{i}\pi} =−\mathrm{12} \\ $$
Commented by hardmath last updated on 28/Jun/24
![thank you dear professors](https://www.tinkutara.com/question/Q208947.png)
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professors} \\ $$