Integral Powers of a Complex Number

Integral power of a complex number is also a complex number. In other words any integral power of a complex number can be expressed in the form of A + iB, where A and B are real.

If z is any complex number, then positive integral powers of z are defined as z\(^{1}\) = a, z\(^{2}\) = z ∙ z, z\(^{3}\) = z\(^{2}\) ∙ z, z\(^{4}\) = z\(^{3}\) ∙ z and so on.

If z is any non-zero complex number, then negative integral powers of z are defined as:

z\(^{-1}\) = \(\frac{1}{z}\), z\(^{-2}\) = \(\frac{1}{z^{2}}\), z\(^{-3}\) = \(\frac{1}{z^{3}}\), etc.

If z ≠ 0, then z\(^{0}\) = 1.

Integral Power of:

Any integral power of i is i or, (-1) or 1.

Integral power of i are defined as:

i\(^{0}\) = 1, i\(^{1}\) = i, i\(^{2}\) = -1,

i\(^{3}\) = i\(^{2}\) ∙ i = (-1)i = -i,

i\(^{4}\) = (i\(^{2}\))\(^{2}\) = (-1)\(^{2}\) = 1,

i\(^{5}\) = i\(^{4}\) ∙ i = 1 ∙ i = i,

i\(^{6}\) = i\(^{4}\) ∙ i\(^{2}\) = 1 ∙ (-1) = -1, and so on.

i\(^{-1}\) = \(\frac{1}{i}\) = \(\frac{1}{i}\) × \(\frac{i}{i}\) = \(\frac{i}{-1}\) = - i

Remember that \(\frac{1}{i}\) = - i

i\(^{-1}\) = \(\frac{1}{i^{2}}\) = \(\frac{1}{-1}\) = -1

i\(^{-3}\) = \(\frac{1}{i^{3}}\) = \(\frac{1}{i^{3}}\) × \(\frac{i}{i}\) = \(\frac{i}{i^{4}}\) = \(\frac{i}{1}\) = i

i\(^{-4}\) = \(\frac{1}{i^{4}}\) = \(\frac{1}{1}\) = 1, and so on.

Note that i\(^{4}\) = 1 and i\(^{-4}\) = 1. It follows that for any integer k,

i\(^{4k}\) = 1, i\(^{4k + 1}\)= i, i\(^{4k + 2}\) = -1, i\(^{4k + 3}\) = - i.

Solved examples on integral powers of a complex number:

1. Express i\(^{109}\) in the form of a + ib.

Solution:

i\(^{109}\)

= i\(^{4 × 27 + 1}\)

= i, [Since, we know that for any integer k, i\(^{4k + 1}\) = i]

= 0 + i, which is the required form of a + ib.

2. Simplify the expression i\(^{35}\) + \(\frac{1}{i^{35}}\) in the form of a + ib.

Solution:

i\(^{35}\) + \(\frac{1}{i^{35}}\)

= i\(^{35}\) + i\(^{-35}\)

= i\(^{4 × 8 + 3}\) + i\(^{4 × (-9) + 1}\)

= 0 + 0

= 0

= 0 + i0, which is the required form of a + ib.

3. Express (1 - i)\(^{4}\) in the standard form a + ib.

Solution:

(1 - i)\(^{4}\)

= [(1 - i)\(^{2}\)]\(^{2}\)

= [1 + i\(^{2}\) - 2i]\(^{2}\)

= (1 + (-1) – 2i)\(^{2}\)

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

= 4i\(^{2}\)

= 4(-1)

= -4

= -4 + i0, which is the required standard form a + ib.

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11 and 12 Grade Math 

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