As we know that \(\sqrt{-1} = i\), where \(i\) is a special symbol first used by Euler .
From above we can conclude that \(\sqrt{-3} = \sqrt{3} i\) and so on. We have \(i^2=-1\) , \(i^3 = -i\), \(i^4=1\), or you can just keep in mind that whenever power of \(i\) is of type \(4m\) ,\(4m+1\), \(4m+2\), \(4m+3\) , where \(m\) is some integer ,then we have
\(i^{4m} = 1\), \(i^{4m+1} = i\), \(i^{4m+2}= -1\) and \(i^{4m+3}=-i\). Lets try some practice questions.
Solve
1). \(i^8\)
2). \(i^4+1-2\)
3). \(i^{10}+i^{12}+i^{11}\)
4). \(2i^{100}-5i^{12}\)
5). \(100i^{101}-2i\)
6). \(8i^8-i^{100}+i^{13}+i^{30}\)
7). \(10i^{22}-10i^{300}-i^{88}\)
8). \((1+i^{32})^2\)
9). \((25+7i)(-3i^23\)
10). \(i^{22}-i^{400}+3i^{22}-82i^{85}\)
Answer
1) 1 2) 0 3) \(-i\) 4) -3 5) \(98i\) 6) \(6+i\)
7) -21 8) 4 9) \(-21+75i\) 10) \(-5-82i\)
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