Lời giải:
•Ta viết $1+i$ dưới dạng lượng giác.
$1+i=\sqrt{2}.(cos\frac{π}{4}+isin\frac{π}{4})$
Theo công thức căn cuả số phức,các căn bậc 10 cuả 1+i là:
$z_k=\sqrt[20]{2}.(cos\frac{\frac{π}{4}+k2π}{10}+isin\frac{\frac{π}{4}+k2π}{10})$ với$k=0,1,2,3,4,5,6,7,8,9$
Vậy $1+i$ có 10 căn bậc 10 là:
$•z_0=\sqrt[20]{2}.(cos\frac{π}{40}+isin\frac{π}{40})$
$•z_1=\sqrt[20]{2}.(cos\frac{9π}{40}+isin\frac{9π}{40})$
$•z_2=\sqrt[20]{2}.(cos\frac{17π}{40}+isin\frac{17π}{40})$
$•z_3=\sqrt[20]{2}.(cos\frac{5π}{8}+isin\frac{5π}{8})$
$•z_4=\sqrt[20]{2}.(cos\frac{33π}{40}+isin\frac{33π}{40})$
$•z_5=\sqrt[20]{2}.(cos\frac{41π}{40}+isin\frac{41π}{40})$
$•z_6=\sqrt[20]{2}.(cos\frac{49π}{40}+isin\frac{49π}{40})$
$•z_7=\sqrt[20]{2}.(cos\frac{57π}{40}+isin\frac{57π}{40})$
$•z_8=\sqrt[20]{2}.(cos\frac{13π}{8}+isin\frac{13π}{8})$
$•z_9=\sqrt[20]{2}.(cos\frac{73π}{40}+isin\frac{73π}{40})$
•Ta có:
$9^{4^{2^{7^{5^{3^{3^{9^{10^{4n^{11}}}}}}}}}}=9122001$
$⇔9^{4^{2^{7^{5^{3^{3^{9^{10^{4n^{11}}}}}}}}}}=9^{4^{2^{7^{5^{3^{3^{9^{10^{log_{10}(log_9(log_3(log_3(log_5(log_7(log_2(log_4(log_99122001))))))))}}}}}}}}}$
$⇔n=\frac{\sqrt[11]{log_{10}(log_9(log_3(log_3(log_5(log_7(log_2(log_4(log_99122001))))))))}}{4}$
