# Khan/khan-exercises

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 Imaginary unit powers
randRange(1, 4) ['1', 'i', '-1', '-i'][EXP % 4]

\large{i ^ {EXP} = ?}

SOLUTION
• 1
• i
• -1
• -i

Anything to the first power is the number itself.

The most important property of the imaginary unit i is that \blue{i ^ 2} = \pink{-1}.

\qquad \begin{eqnarray} i ^ 3 &=& (\pink{i ^ 2}) \cdot i \\ &=& (\blue{-1}) \cdot i \\ &=& -i \end{eqnarray}

\qquad \begin{eqnarray} i ^ 4 &=& (\pink{i ^ 2}) ^ 2 \\ &=& (\blue{-1}) ^ 2 \\ &=& 1 \end{eqnarray}

\qquad i ^ EXP = SOLUTION

4 + randRange(1, 30) ['1', 'i', '-1', '-i'][EXP % 4] floor(EXP / 4) EXP % 4 i18n._("remainder")

\large{i ^ {EXP} = ?}

SOLUTION
• 1
• i
• -1
• -i

The most important property of the imaginary unit i is:
\qquad \blue{i ^ 2} = \pink{-1}

Therefore:
\qquad i ^ 4 = (\blue{i ^ 2}) ^ 2 = (\pink{-1}) ^ 2 = 1

So, we can simplify the expression by rewriting it in terms of i^4.

Because EXP \div \blue{4} = \green{WHOLES} \text{ REMAINDER_TEXT } \red{REMAINDER},

\qquad \begin{eqnarray} i ^ {EXP} &=& (i^\blue{4})^\green{WHOLES} \cdot i^\red{REMAINDER} \\ &=& (1)^\green{WHOLES} \cdot i^\red{REMAINDER} \\ &=& i^\red{REMAINDER} 1 \end{eqnarray}

Anything to the first power is the number itself.
\qquad i ^ \red{1} = i

As stated above, \blue{i ^ 2} = \pink{-1}.

\qquad \begin{eqnarray} i ^ \red{3} &=& (\blue{i ^ 2}) \cdot i \\ &=& (\pink{-1}) \cdot i \\ &=& -i \end{eqnarray}

i ^ {EXP} = i ^ {REMAINDER} = SOLUTION