Problem type

All the n-th roots of a complex number

What it solves: Find every square root, cube root or higher root of a complex number. There are always exactly n of them, equally spaced round a circle, and each one is checked by raising it back to the power.
Subject: Complex Numbers
Topic: Length and angle
Also called: nth roots, cube root of a complex number, roots of a complex number
Free example: Read the complete worked example

Femtobytes

All 2 square roots of -3 + 2i, step by step

femtobytes.com/solve/complex-roots/-3_1_2_1_2

The problem

Find all the square roots of this complex number: every $w$ that satisfies the equation.

w^{2} = -3 + 2 i

The result

w_k = \sqrt{\sqrt{13}} \left( \cos \frac{\theta + 2k\pi}{2} + i \sin \frac{\theta + 2k\pi}{2} \right), \quad k = 0, 1

Here $\theta \approx 2.5536$ . The first root is $w_0 \approx 0.5503 + 1.8174 i$ ; the full list is in the table below.

checked 15 lines of this solution were evaluated with exact arithmetic before it was shown to you. A solution with a false line is never served.

Picture 1 of 2Step 1: See where the number lives

The number z drawn as an arrow from the origin. Its length is the modulus r and the angle it makes with the positive real axis is the argument, written with the Greek letter theta. Here

r = \sqrt{13}

and

\theta = 2.5536

See where the number lives

Why this step

A complex number is two real numbers travelling together, so it cannot sit on a number line; it needs a plane. Rectangular form $a + bi$ says "go $a$ across and $b$ up". Polar form describes the SAME point as "face this direction, walk this far". The reason to bother: when complex numbers are multiplied, their lengths multiply and their directions add. Rectangular form hides that completely; polar form makes it visible.

Here $z = -3 + 2 i$ , so the real part is $a = -3$ and the imaginary part is $b = 2$ . The symbol $i$ is the imaginary unit, the number with $i^2 = -1$ . On the plane, $z$ is the point $(a, b)$ : $a$ is read on the horizontal (real) axis and $b$ on the vertical (imaginary) axis.

The real part is negative and the imaginary part positive, so $z$ lies in the second quadrant: to the left and above. Its angle will be between $\frac{\pi}{2}$ and $\pi$ .

The number z drawn as an arrow from the origin. Its length is the modulus r and the angle it makes with the positive real axis is the argument, written with the Greek letter theta. Here

r = \sqrt{13}

and

\theta = 2.5536

Find the modulus: how far z is from the origin

Why this step

The arrow to $z$ is the long side of a right-angled triangle whose other sides are $\left| a \right|$ across and $\left| b \right|$ up. Pythagoras gives its length. We square $a$ and $b$ first, which also throws away their signs: a distance cannot care whether we went left or right.

\displaystyle r

=

\displaystyle \sqrt{\left( -3 \right)^{2} + 2^{2}}

The modulus

r = \left| z \right| = \sqrt{a^2 + b^2}

, by Pythagoras. checked

=

\displaystyle \sqrt{9 + 4}

Square each part. Squaring a square root undoes it, for example

\left( \sqrt{3} \right)^2 = 3

, and the square of a negative number is positive. checked

=

\displaystyle \sqrt{13}

Add. checked

The same arrow, with the right-angled triangle it is the long side of. The side across is the real part and the side up is the imaginary part, so the modulus is the length Pythagoras gives from those two.

All the n-th roots of a complex number

See where the number lives

Find the modulus: how far z is from the origin

Find the argument: which way z points

Write the polar form, and check it by turning it back

Turn "find the roots" into an equation about lengths and angles

Find the length of the roots

Find the angles, one for each k

Write out the roots

Check every root by raising it back to the power