Question

Use the Binomial theorem, write the number (1 + 2i)^12 in the form a + ib.

Answer 1

$(1+2i)^{12}= \dbinom{12}{0}1^{12}(2i)^0 + \dbinom{12}{1}1^{11}(2i)^1 + \dbinom{12}{2}1^{10}(2i)^2 + \dbinom{12}{3} 1^9(2i)^3 + \dbinom{12}{4}1^8(2i)^4 + \dbinom{12}{5}1^7(2i)^5 + \dbinom{12}{6}1^6(2i)^6 + \dbinom{12}{7}1^5(2i)^7 + \dbinom{12}{8} 1^4(2i)^8 +\dbinom{12}{9}1^3(2i)^9 + \dbinom{12}{10}1^2(2i)^{10} + \dbinom{12}{11}1^1(2i)^{11} + \dbinom{12}{12}1^0(2i)^{12}=\\ =1\cdot 1+12\cdot 2i-66\cdot 4-220\cdot 8i+495\cdot 16+792\cdot 32i-924\cdot 64-792\cdot 128i+495\cdot 256+220\cdot 512i-66\cdot 1024-12\cdot 2048i+1\cdot 4096=\\ =1+24i-264-1760i+7920+25344i-59136–101376i+126720+112640i-67584-24576i+4096=\\ = 11753-10296i$

Answer: $(1+2i)^{12}=11753-10296i$