2 cis 30° times 3 cis 45° really is 6 cis 75°: multiply the moduli, add the arguments. Do the same to the sum and you get 5 cis 75°. The sum is 4.96 cis 39.01°, and both parts are wrong.
The product follows a rule. The sum needs the parallelogram, which means going back to Cartesian form.
| Form | Written | 2 cis 30° looks like |
|---|---|---|
| Cartesian | z = a + bi | 1.732 + i |
| Polar | z = r(cosθ + i sinθ) = r cis θ | 2 cis 30° |
| Euler | z = reiθ | 2eiπ/6 |
Euler form needs the argument in radians, always, because the exponent of e cannot carry a degree symbol. 30° is π/6, which is 0.524. Writing 2e30i is a different number entirely: 30 radians is more than four complete turns.
Converting from polar to Cartesian is substitution:
2 cis 30° = 2cos30° + 2i sin30° = 1.732 + i
and back the other way is the modulus and argument work from 1.12, with the quadrant check that goes with it:
r = √(a² + b²), θ = arctan(b/a) placed in the right quadrant
Nothing new, and the quadrant trap from 1.12 is still live here.
For z₁ = r₁ cis θ₁ and z₂ = r₂ cis θ₂:
z₁z₂ = r₁r₂ cis(θ₁ + θ₂) and z₁/z₂ = (r₁/r₂) cis(θ₁ − θ₂)
so with 2 cis 30° and 3 cis 45°:
In Euler form this is just the exponent law: r₁eiθ₁ × r₂eiθ₂ = r₁r₂ei(θ₁+θ₂). That is why the rule looks arbitrary in polar form and obvious in Euler form, and it is the reason to learn both.
Geometrically, multiplying by a complex number is a rotation and a stretch. Multiply anything by 3 cis 45° and it turns 45° anticlockwise and grows by a factor of 3. The clearest case is multiplying by i, which is 1 cis 90°:
(3 + 4i) × i = −4 + 3i
Modulus still 5, argument moved from 53.13° to 143.13°. A quarter turn and nothing else, which is as clean a demonstration of the rule as there is.
Nothing combines two polar forms into the polar form of their sum. To add, convert to Cartesian, add, and convert back:
Now compare the number you get from treating addition like multiplication, 5 cis 75°, which in Cartesian form is 1.294 + 4.830i. It is not close. The real part is out by 2.6 and the imaginary part by 1.7.
Notice also that 4.96 is less than 5. It has to be: adding two arrows that point in different directions gives something shorter than their lengths added, which is the triangle inequality,
|z₁ + z₂| ≤ |z₁| + |z₂|
with equality only when the two arguments are equal. So a "sum" with modulus exactly 5 is only possible if both numbers point the same way, and these do not.
Choose the form that suits the operation.
Euler's identity. Put r = 1 and θ = π into Euler form:
eiπ = cosπ + i sinπ = −1, so eiπ + 1 = 0
It is not examinable as a fact to recall, but it is one line from the form you have just learned, and it is the shortest way to check you have the form the right way round. If your version of Euler form does not give −1 at θ = π, it is wrong.
Both machines convert between the forms directly and will multiply in either, so the arithmetic is cheap. The thinking that remains is choosing the form, and no machine does that for you.
When you may use it. Analysis Paper 1 is non-calculator, and polar-form questions there are built to be exact: arguments of π/6, π/4 and π/3, and moduli like 2 and √2. Paper 2 and Paper 3 allow it, and that is where 4.96 and 39.01° are acceptable answers.
The mark people lose. Adding the moduli and the arguments. It is the right rule for the wrong operation, and it gives 5 cis 75° where the answer is 4.96 cis 39.01°. The check that catches it in one line: a sum's modulus can never exceed the sum of the moduli, so if your answer's modulus is exactly 5 when the parts are 2 and 3, the two numbers must point the same way, and these are 15° apart. Also watch the mode: in radians, 2∠30 means 30 radians.
1. Find the modulus of (2 cis 30°)(3 cis 45°).
2. Find its argument, in degrees.
3. Now find the modulus of their sum, to 2 decimal places.
4. And the argument of the sum, in degrees to 2 decimal places.
5. Why can the modulus of the sum not be 5?
1 markThe right form chosen for the operation.
1 markThe conversion, where one is needed.
1 markThe arithmetic.
1 markThe answer in the form the question asked for.
The last mark catches more people than the arithmetic does. If a question says "give your answer in the form a + bi", a polar answer scores nothing for it, however right the number is.
These pages are free and stay free, but they are general and your IA is not. Send me your research question, or whatever exists so far, and I will tell you in writing whether the topic has a ceiling on it, where the marks are going, and what to change first. That costs nothing and it comes back within 24 hours.
Written by a serving IB Diploma and Career-related Programme Coordinator and Head of Mathematics, who reads internal assessments across every subject group every year. If you then want the whole draft reviewed properly against all five criteria, that is the paid one, and it is refunded if it does not name at least three specific things to fix.
Already have a full draft? Have the whole thing reviewed against all five criteria, $99.