Powers of ii

In complex numbers, ii is defined as the imaginary unit, satisfying:

i=1i = \sqrt{-1}

This means i2=1i^2 = -1, which gives rise to an interesting pattern when we compute higher powers of ii.

The Pattern

Let's compute the first few powers of ii:

i1=ii^1 = i

i2=(1)2=1i^2 = {(\sqrt{-1})}^2 = -1

i3=i2i=1i=ii^3 = i^2 \cdot i = -1 \cdot i = -i

i4=i2i2=(1)(1)=1i^4 = i^2 \cdot i^2 = (-1) \cdot (-1) = 1

After that, the powers repeat every 4 steps:

i5=ii^5 = i

i6=1i^6 = -1

i7=ii^7 = -i

i8=1i^8 = 1

General Rule

The powers of ii cycle every 4. So to simplify ini^n, we reduce the exponent modulo 4:

in=inmod4i^n = i^{n \bmod 4}

And since the pattern is:

i1=ii2=1i3=ii0=1\begin{aligned} i^1 &= i \\ i^2 &= -1 \\ i^3 &= -i \\ i^0 &= 1 \end{aligned}

For example, i23i^{23} simplifies to i3=ii^3 = -i because 23mod4=323 \bmod 4 = 3.

Let's Practice

Simplify the following:

i5i^{5}

Simplify the following:

i5-i^{5}

Simplify the following:

i5i^{5}

Simplify the following:

(i)2(-i)^{2}

Simplify the following:

i2i^{2}


End of Lesson

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Addition of Complex Numbers
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Coverting Square Roots to Iota