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.

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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.

Test yourself — a quick 14-question Complex Numbers recall quiz, instant score.

Concept 1 of 2: Powers of i (the period-4 cycle)

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 $1\pm i$ reduce to powers of $i$ : $(1+i)^2=2i$ , $(1-i)^2=-2i$ , $\dfrac{1+i}{1-i}=i$ , $\dfrac{1-i}{1+i}=-i$ . So $\Big(\dfrac{1+i}{1-i}\Big)^n=i^n$ equals 1 first at $n=4$ and $-1$ first at $n=2$ ; $\Big(\dfrac{1-i}{1+i}\Big)^{n^2}=(-i)^{n^2}=1$ needs $4\mid n^2$ , i.e. $n$ even — smallest $n=2$ ; $(1+i)^{2n}=(2i)^n=2^n i^n$ , real iff $n$ is even.

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

The same idea in a real exam question:

NDA · Apr 2017 · Q3Easy

Example 1 · Complex Numbers · Powers and Roots

The value of

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

, where

i = \sqrt{-1}

, is

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

Concept 2 of 2: De Moivre's theorem and roots

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 ( $x^2-y^2=a$ , $2xy=b$ ).

Closed form of the square root: $\sqrt{a+ib}=\pm\Big(\sqrt{\tfrac{|z|+a}{2}}+i\,\operatorname{sgn}(b)\sqrt{\tfrac{|z|-a}{2}}\Big)$ with $|z|=\sqrt{a^2+b^2}$ . Hence $\big(\sqrt{a+ib}+\sqrt{a-ib}\big)^2=2a+2\sqrt{a^2+b^2}$ (the cross term is $2|z|$ ) — e.g. $\sqrt{3+4i}+\sqrt{3-4i}=\pm\sqrt{16}=\pm4$ .
Evaluating a polynomial at a complex number — the minimal-polynomial trick: from $x=4+i$ form $(x-4)^2=-1$ , i.e. $x^2-8x+17=0$ ; divide the given polynomial by $x^2-8x+17$ and evaluate only the REMAINDER at $x=4+i$ . Any polynomial with real coefficients that vanishes at $a+ib$ is divisible by $x^2-2ax+(a^2+b^2)$ .

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

The same idea in a real exam question:

NDA · Sep 2023 · Q12Hard

Example 2 · Complex Numbers · Powers and Roots

Which one of the following is a square root of

\sqrt{-\sqrt{-1}}

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)

Test yourself on Complex Numbers

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