Geometry: An Interactive Journey to Mastery

D௘ͽ ii i
E௘ͽ iii
F௘ͽ u ii i

D௘ͽ )URPͼ௘a + ib௘ͽͼ௘i௘ͽ ZHVHHWKDWZHQHHG
ͼ௘aíb௘ͽiͼ௘a + 4b௘ͽ
This suggests solving
4 aíb = 1
3 a + 4b = 0.
The second equation gives ba ^34.
6XEVWLWXWLQJLQWRWKH¿UVW\LHOGV41,aa 49 giving ab 2543 ,so 25.
It seems that 4 3^143 i 25 25 i.
E௘ͽ )URPͼ௘íi௘ͽi = 1, it seems that^1 i i.
F௘ͽ )ROORZLQJWKHVDPHOLQHRIUHDVRQLQJDVSDUWD௘ͽLWVHHPVWKDWpiq^1 pq pq22 22pqi.

From i^2 íZHVHHWKDWi^4 ͼ௘i^2 ௘ͽ^2 ͼ௘í௘ͽ^2 = 1. Thus, i^400 ͼ௘i^4 ௘ͽ^100 = 1^100 = 1. Consequently,
iiii^403 ^400211 ii.

Geometry: An Interactive Journey to Mastery

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