Femtobytes
Problem type

Polar form of a complex number

What it solves
Convert a complex number a + bi to polar form: find its modulus (length) and its principal argument (direction), exactly where the angle is a special one, and check the result by converting it back.
Subject
Complex Numbers
Topic
Length and angle
Also called
modulus and argument, polar form, argument of a complex number, Argand
Your problem

Write this complex number in polar form.

z=−2+8iz = -2 + 8 i
Invent a new one:
The problem

Write this complex number in polar form.

z=−2+8iz = -2 + 8 i
The result
z=−2+8i=217(cos⁡1.8158+isin⁡1.8158)z = -2 + 8 i = 2 \sqrt{17} \left( \cos 1.8158 + i \sin 1.8158 \right)

The modulus is r=217r = 2 \sqrt{17} and the principal argument is θ≈1.8158\theta \approx 1.8158 radians.

checked 9 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=217r = 2 \sqrt{17} and θ=1.8158\theta = 1.8158.
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=−2+8iz = -2 + 8 i, so the real part is a=−2a = -2 and the imaginary part is b=8b = 8. 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==(−2)2+82\displaystyle \sqrt{\left( -2 \right)^{2} + 8^{2}}
The modulus r=∣z∣=a2+b2r = \left| z \right| = \sqrt{a^2 + b^2}, by Pythagoras. checked
==4+64\displaystyle \sqrt{4 + 64}
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
==68\displaystyle \sqrt{68}
Add. checked
==217\displaystyle 2 \sqrt{17}
Take the square factor out of the root: 68=4⋅1768 = 4 \cdot 17, and 4=2\sqrt{4} = 2. checked
3

Find the argument: which way z points

4

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

5

Every other angle that names the same direction