NDA Maths · Complex Numbers

Powers of i, De Moivre & Roots

The four-step cycle of powers of i, De Moivre's theorem for raising complex numbers to powers, and finding square/nth roots of a complex number.

All Complex Numbers notes NDA Maths strategy

Why this matters

Powers-of-i questions are fast marks once you use the period-4 cycle, and De Moivre turns an ugly (1+i)ⁿ computation into one angle multiplication.

Concept 1 of 2

Powers of i (the period-4 cycle)

Intuition

Powers of

i

repeat every four:

i,-1,-i,1,

then back to

i

. So any

i^n

is decided by

n \bmod 4

, and any block of four consecutive powers sums to zero — which collapses long sums instantly.

Definition

$i^1=i,\;i^2=-1,\;i^3=-i,\;i^4=1$ ; thereafter $i^n=i^{\,n\bmod 4}$ . Block sum: $i^k+i^{k+1}+i^{k+2}+i^{k+3}=0$ for any $k$ . So $\sum$ of $i^n$ over a full set of consecutive 4 is 0 — only the leftover terms survive.

Powers of i

i^2=-1 \qquad i^3=-i \qquad i^4=1 \qquad i^{4k+r}=i^r \qquad i^k+i^{k+1}+i^{k+2}+i^{k+3}=0

Worked example

Evaluate

i^{1000}+i^{1001}+i^{1002}+i^{1003}

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Complex NumbersEASY

The value of

i^{2n} + i^{2n+1} + i^{2n+2} + i^{2n+3}

, where

i = \sqrt{-1}

, is

[Q3 · Apr · 2017]

Reduce the exponent of $i$ mod 4 — and watch the remainder

i^n=i^{\,n\bmod 4}

: take the exponent mod 4, not the whole number mod something else. Remainder

0\to1

1\to i

2\to-1

3\to-i

. So

i^{102}

102\bmod4=2\Rightarrow i^{102}=-1

(a frequent error is reading remainder 2 as

i

instead of

-1

Drill 9 more on powers of i (the period-4 cycle)

Concept 2 of 2

De Moivre's theorem and roots

Intuition

To raise a complex number to a power, put it in polar form and multiply the angle:

(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta

. Running it backwards gives the

n

nth-roots, equally spaced around a circle. Square roots of

a+ib

can also be found by solving

(x+iy)^2=a+ib

Definition

De Moivre: $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$ (also for the modulus: $z^n=r^n e^{in\theta}$ ). nth roots of $re^{i\theta}$ : $r^{1/n}e^{i(\theta+2k\pi)/n}$ , $k=0,\ldots,n-1$ — $n$ points on a circle of radius $r^{1/n}$ . Square root of $a+ib$ : set $(x+iy)^2=a+ib$ , match parts ( $x^2-y^2=a$ , $2xy=b$ ).

De Moivre's theorem and nth roots

(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta \qquad z^n=r^n e^{in\theta} \qquad z^{1/n}=r^{1/n}e^{i(\theta+2k\pi)/n},\ k=0,\ldots,n-1

Worked example

Use De Moivre to find

(1+i)^4

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Complex NumbersHARD

Which one of the following is a square root of

\sqrt{-\sqrt{-1}}

[Q12 · Sep · 2023]

Drill 4 more on de moivre's theorem and roots

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Powers of i (the period-4 cycle)
Powers of i
$i^2=-1 \qquad i^3=-i \qquad i^4=1 \qquad i^{4k+r}=i^r \qquad i^k+i^{k+1}+i^{k+2}+i^{k+3}=0$
De Moivre's theorem and roots
De Moivre's theorem and nth roots
$(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta \qquad z^n=r^n e^{in\theta} \qquad z^{1/n}=r^{1/n}e^{i(\theta+2k\pi)/n},\ k=0,\ldots,n-1$

Watch out for (1)

Reduce the exponent of $i$ mod 4 — and watch the remainder
→ Powers of i (the period-4 cycle)

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