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=−5−5iz = -5 - 5 i
Invent a new one:
The problem

Write this complex number in polar form.

z=−5−5iz = -5 - 5 i
The result
z=−5−5i=52(cos⁡(−3π4)+isin⁡(−3π4))z = -5 - 5 i = 5 \sqrt{2} \left( \cos \left( -\frac{3 \pi}{4} \right) + i \sin \left( -\frac{3 \pi}{4} \right) \right)

The modulus is r=52r = 5 \sqrt{2} and the principal argument is θ=−3π4\theta = -\frac{3 \pi}{4}.

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
22i-2-2i44i-4-4i66i-6-6i88i-8-8ireal 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=52r = 5 \sqrt{2} and θ=−3π4\theta = -\frac{3 \pi}{4}.
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=−5−5iz = -5 - 5 i, so the real part is a=−5a = -5 and the imaginary part is b=−5b = -5. 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.

Both parts are negative, so zz lies in the third quadrant: to the left and below. Its principal angle will be between −π-\pi and −π2-\frac{\pi}{2}.

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==(−5)2+(−5)2\displaystyle \sqrt{\left( -5 \right)^{2} + \left( -5 \right)^{2}}
The modulus r=∣z∣=a2+b2r = \left| z \right| = \sqrt{a^2 + b^2}, by Pythagoras. checked
==25+25\displaystyle \sqrt{25 + 25}
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
==50\displaystyle \sqrt{50}
Add. checked
==52\displaystyle 5 \sqrt{2}
Take the square factor out of the root: 50=25⋅250 = 25 \cdot 2, and 25=5\sqrt{25} = 5. 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