Differential-Equations · Unit 6 · Video 1 · Interactive Practice

Complex Division, Polar Form, and the Definition of e^(iθ)

IKey Formulas

Formula	Name	What it says
$z = a + bi,\quad \bar z = a - bi,\quad z\bar z = a^2 + b^2$	Complex conjugate	Mirror image across the real axis; a number times its conjugate is real
$\dfrac{w}{z} = \dfrac{w\,\bar z}{z\,\bar z}$	Division	Multiply top and bottom by the conjugate of the denominator, then split into real and imaginary parts
$a + bi = r(\cos\theta + i\sin\theta),\quad r = \sqrt{a^2 + b^2}$	Polar form	$a = r\cos\theta$ , $b = r\sin\theta$ , with $\theta$ measured from the positive real axis
$e^{i\theta} := \cos\theta + i\sin\theta$	Euler's definition	Gives $e$ to an imaginary power a meaning it did not have before

Key Insight: Every complex number is its distance from the origin times $\cos\theta + i\sin\theta$ , an expression that depends only on the angle. Calling it $e^{i\theta}$ is a definition; it earns the name only if it obeys the law of exponents and solves $y' = iy$ , $y(0) = 1$ .

IIDividing by the Conjugate

Multiplying top and bottom by $\bar z$ makes the denominator the real number $|z|^2$ , so $w/z$ splits into $a + bi$ .

IIIPolar Form: Distance Times an Angle Factor

Any point $a + bi$ is $r$ times the unit-circle point $\cos\theta + i\sin\theta$ , which depends only on $\theta$ .

IVWhat Makes Something an Exponential

Every $y = Ce^{at}$ solves $y' = ay$ ; only the initial condition $y(0) = 1$ singles out $e^{at}$ itself.

💡 Euler's $e^{i\theta}$ must pass the same test with $a$ replaced by $i$ — $\frac{d}{dt}e^{it} = i\,e^{it}$ with $e^{i\cdot 0} = 1$ — and obey $e^{i\theta_1}e^{i\theta_2} = e^{i(\theta_1+\theta_2)}$ , just as $2^3\cdot 2^4 = 8\cdot 16 = 128 = 2^7$ .

VQuiz Questions

Problem 1 · Divide Two Complex Numbers

Given: $\dfrac{3 + 2i}{1 + i}$ — find it in the form $a + bi$ .

✅ Correct! The conjugate

1 - i

turns the denominator into

1^2 + 1^2 = 2

, and the numerator becomes

5 - i

❌ Check $i^2$ . In the numerator,

2i\cdot(-i) = -2i^2 = +2

, not

-2

: the real part is

3 + 2 = 5

❌ Division is not done part by part.

(3+2i)/(1+i)

is not

3/1 + (2/1)i

; check by multiplying back:

(3+2i)(1+i) = 1 + 5i \ne 3 + 2i

❌ You multiplied the top but not the bottom.

(1-i)/(1-i)

is the number one only as a whole fraction; the denominator becomes

(1+i)(1-i) = 2

❌ Not quite. Multiply top and bottom by

1 - i

, the conjugate of the denominator.

Show solution

Multiply by the conjugate of the denominator, $\dfrac{1 - i}{1 - i} = 1$ :

\frac{3 + 2i}{1 + i}\cdot\frac{1 - i}{1 - i} = \frac{(3 + 2i)(1 - i)}{(1 + i)(1 - i)}

Denominator: $(1 + i)(1 - i) = 1^2 + 1^2 = 2$ .

Numerator, real part then imaginary part: real $= 3\cdot 1 + (2i)(-i) = 3 + 2 = 5$ ; imaginary $= 3\cdot(-i) + 2i\cdot 1 = -i$ . So the numerator is $5 - i$ .

\frac{3 + 2i}{1 + i} = \frac{5 - i}{2} = \frac{5}{2} - \frac{1}{2}i

Check: $\left(\tfrac{5}{2} - \tfrac{1}{2}i\right)(1 + i) = \tfrac{5}{2} + \tfrac{5}{2}i - \tfrac{1}{2}i + \tfrac{1}{2} = 3 + 2i$ ✓

Problem 2 · Polar Form in the Third Quadrant

Given: $z = -1 - \sqrt{3}\,i$ — find its polar form $r(\cos\theta + i\sin\theta)$ .

✅ Correct!

r = 2

, and the point lies in the third quadrant,

\tfrac{\pi}{3}

past the negative real axis.

❌ Wrong quadrant.

\tan\theta = \sqrt{3}

holds at

\tfrac{\pi}{3}

AND at

\tfrac{4\pi}{3}

;

2(\cos\tfrac{\pi}{3} + i\sin\tfrac{\pi}{3}) = 1 + \sqrt{3}\,i

, the opposite point.

❌ $r$ is a distance.

a^2 + b^2 = 4

r^2

; the distance from the origin is

r = \sqrt{4} = 2

❌ Check the real part.

\cos\tfrac{5\pi}{3} = \tfrac{1}{2}

, so this is

1 - \sqrt{3}\,i

, in the fourth quadrant;

z

has a negative real part.

❌ Not quite. Find

r = \sqrt{a^2 + b^2}

, then the angle whose cosine is

a/r

and whose sine is

b/r

Show solution

Distance: $r = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2$ .

Angle: we need $\cos\theta = a/r = -\tfrac{1}{2}$ and $\sin\theta = b/r = -\tfrac{\sqrt{3}}{2}$ . Both negative means the third quadrant, so $\theta = \pi + \tfrac{\pi}{3} = \tfrac{4\pi}{3}$ (equivalently $-\tfrac{2\pi}{3}$ ).

-1 - \sqrt{3}\,i = 2\left(\cos\tfrac{4\pi}{3} + i\sin\tfrac{4\pi}{3}\right) = 2e^{i\,4\pi/3}

The shortcut $\theta = \arctan(b/a) = \arctan\sqrt{3} = \tfrac{\pi}{3}$ lands on $1 + \sqrt{3}\,i$ instead: $b/a$ cannot tell a point from its opposite.

Problem 3 · Divide, Then Go Polar

Given: $w = \dfrac{1 + 7i}{4 + 3i}$ — find $w$ in the form $a + bi$ , then its polar form.

What is w in the form a + bi?

What is its polar form?

✅ Correct!

w = 1 + i = \sqrt{2}\left(\cos\tfrac{\pi}{4} + i\sin\tfrac{\pi}{4}\right) = \sqrt{2}\,e^{i\pi/4}

❌ Check $i^2$ .

(7i)(-3i) = -21i^2 = +21

, so the real part of the numerator is

4 + 21 = 25

, not

4 - 21

❌ The conjugate product is a sum of squares.

(4 + 3i)(4 - 3i) = 4^2 + 3^2 = 25

, not

4^2 - 3^2 = 7

❌ Division is not done part by part. Multiply top and bottom by

4 - 3i

instead of dividing

1

4

and

7

3

❌ Check the quotient. Multiply top and bottom by

4 - 3i

❌ Missing the distance. This is the unit-circle point;

w

r = \sqrt{1^2 + 1^2} = \sqrt{2}

times it.

❌ $r$ is a distance.

a^2 + b^2 = 2

r^2

;

r = \sqrt{2}

❌ Check the polar form.

1 + i

lies in the first quadrant, on the line

b = a

, at distance

\sqrt{2}

Show solution

Step 1: divide. Multiply top and bottom by $4 - 3i$ .

Denominator: $(4 + 3i)(4 - 3i) = 4^2 + 3^2 = 25$ .

Numerator: real part $1\cdot 4 + (7i)(-3i) = 4 + 21 = 25$ ; imaginary part $1\cdot(-3i) + 7i\cdot 4 = 25i$ . So

w = \frac{25 + 25i}{25} = 1 + i

Step 2: polar form. $r = \sqrt{1^2 + 1^2} = \sqrt{2}$ ; $\cos\theta = \sin\theta = \tfrac{1}{\sqrt{2}}$ gives $\theta = \tfrac{\pi}{4}$ .

w = \sqrt{2}\left(\cos\tfrac{\pi}{4} + i\sin\tfrac{\pi}{4}\right) = \sqrt{2}\,e^{i\pi/4}

Problem 4 · The Reciprocal of the Angle Factor

Given: the definition $e^{i\theta} := \cos\theta + i\sin\theta$ — find $\dfrac{1}{e^{i\theta}}$ in the form $a + bi$ .

✅ Correct! The conjugate over

\cos^2\theta + \sin^2\theta = 1

— and

\cos\theta - i\sin\theta = \cos(-\theta) + i\sin(-\theta) = e^{-i\theta}

, just what an exponential should give.

❌ Division is not done part by part.

1/(a + bi) \ne 1/a + 1/(bi)

; multiply top and bottom by the conjugate

\cos\theta - i\sin\theta

❌ The conjugate product is a sum of squares.

(\cos\theta + i\sin\theta)(\cos\theta - i\sin\theta) = \cos^2\theta + \sin^2\theta = 1

, not a difference.

❌ That is $e^{i\theta}$ itself. Its product with

e^{i\theta}

\cos 2\theta + i\sin 2\theta

, which is not

1

in general.

❌ Not quite. Treat

1/(\cos\theta + i\sin\theta)

as a complex division and use the conjugate.

Show solution

Multiply top and bottom by the conjugate $\cos\theta - i\sin\theta$ :

\frac{1}{\cos\theta + i\sin\theta}\cdot\frac{\cos\theta - i\sin\theta}{\cos\theta - i\sin\theta} = \frac{\cos\theta - i\sin\theta}{\cos^2\theta + \sin^2\theta} = \cos\theta - i\sin\theta

Since cosine is even and sine is odd, $\cos\theta - i\sin\theta = \cos(-\theta) + i\sin(-\theta)$ , which by the definition is $e^{-i\theta}$ . So $1/e^{i\theta} = e^{-i\theta}$ — consistent with $a^{-x} = 1/a^{x}$ for ordinary exponentials. It is also the mirror image $\overline{e^{i\theta}}$ : the unit-circle point reflected across the real axis.

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