Femtobytes
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
Your problem

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

w2=−4+7iw^{2} = -4 + 7 i
Invent a new one:
The problem

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

w2=−4+7iw^{2} = -4 + 7 i
The result
wk=65(cos⁡θ+2kπ2+isin⁡θ+2kπ2),k=0,1w_k = \sqrt{\sqrt{65}} \left( \cos \frac{\theta + 2k\pi}{2} + i \sin \frac{\theta + 2k\pi}{2} \right), \quad k = 0, 1

Here θ≈2.0899\theta \approx 2.0899. The first root is w0≈1.4252+2.4558iw_0 \approx 1.4252 + 2.4558 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
55i-5-5i1010i-10-10ireal axisimaginary axisθzr
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=65r = \sqrt{65} and θ=2.0899\theta = 2.0899.
1

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+bia + bi says "go aa across and bb 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=−4+7iz = -4 + 7 i, so the real part is a=−4a = -4 and the imaginary part is b=7b = 7. The symbol ii is the imaginary unit, the number with i2=−1i^2 = -1. On the plane, zz is the point (a,b)(a, b): aa is read on the horizontal (real) axis and bb on the vertical (imaginary) axis.

The real part is negative and the imaginary part positive, so zz lies in the second quadrant: to the left and above. Its angle will be between π2\frac{\pi}{2} and π\pi.

2

Find the modulus: how far z is from the origin

Why this step

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

r\displaystyle r==(−4)2+72\displaystyle \sqrt{\left( -4 \right)^{2} + 7^{2}}
The modulus r=∣z∣=a2+b2r = \left| z \right| = \sqrt{a^2 + b^2}, by Pythagoras. checked
==16+49\displaystyle \sqrt{16 + 49}
Square each part. Squaring a square root undoes it, for example (3)2=3\left( \sqrt{3} \right)^2 = 3, and the square of a negative number is positive. checked
==65\displaystyle \sqrt{65}
Add. checked
3

Find the argument: which way z points

4

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

5

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

6

Find the length of the roots

7

Find the angles, one for each k

8

Write out the roots

9

Check every root by raising it back to the power