# 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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