# Convert Complex Numbers to Trigonometric Form

## Convert Complex Numbers to Trigonometric Form - How it Works - Video

### Formulas Formulas:

These are the formulas that we will use to transform a complex number into trigonometric form.

### Example 1 Example 1:

Here we have our number in standard form, z = a + b * i. We must convert it to the form z = r * (cosθ + i * sinθ). We know r = sqrt( a2 + b2) and tanθ = b/a.

If we use the a and b of our complex number to form a triangle. The modulus of z or r or the distance between the origin and our point is the hypotenuse. Now we can use a and b as our legs to find our missing side length.

22 + 42 = c2

sqrt(22 + 42) = c

sqrt(4 + 16) = c

sqrt(20) = c

sqrt(4) * sqrt(5) = c

2 * sqrt(5) = c

4.47 = c

Pythagorean Theorem

Substituted 2 for a and 4 for b

Squared both numbers

Found square root

Converted to Decimal

Now we know c, the hypotenuse, or our r, 2 * sqrt(5). Now, we must solve for our argument or theta, by using tanθ = b/a. Remember b is the opposite side length while a is the adjacent side length.

tanθ = b/a

tanθ = 4/2

tanθ = 2

θ = tan-1(2)

θ = 63.1°

Substituted 2 for a and 4 for b

Divided

Used the inverse tangent on both sides

Calculated

Since we are in Quadrant I, we have the theta that we want. We also have our r, so now we can use z = r * (cosθ + i * sinθ). Our final answer is z = 2sqrt(5) * (cos63.1° + i * sin63.1°).

### Example 2 Example 2:

Here we have our number in standard form, z = a + b * i. We must convert it to the form z = r * (cosθ + i * sinθ). We know r = sqrt( a2 + b2) and tanθ = b/a.

If we use the a and b of our complex number to form a triangle. The modulus of z or r or the distance between the origin and our point is the hypotenuse. Now we can use a and b as our legs to find our missing side length. But before we graph we have simplify the number we have in the question, 2i * (3 + 2i).

2i * (3 + 2i)

2i * 3 + 2i * 2i

6i + 4i2

6i + 4 * (-1)

-4 + 6i

Distributed 2i

Multiplied terms

Simplified i2

Multiplied and moved terms around Here is where trouble starts. The calculator doesn't know which quadrant the triangle is in. This means it doesn't know if a or b is negative. We, the humans, know, so we have to use our drawing to figure out what the calculator is telling us. If we go counterclockwise, the negative direction, Quadrant IV is from 0° to -90°. That means the -56.3° is in Quadrant IV, and gives us a different triangle than our original. To find θ1= we need to add 180° to -56.3°, which is 123.7°. If we use positive numbers for both a and b, we know the angle is 56.3°.

tanθ2= b/a

tanθ2 = 6/4

tanθ2 = 3/2

θ2 = tan-1(3/2)

θ2 = 56.3°

Substitute 2 for a and 4 for b

Divide

Inverse tangent on both sides

Calculate

Since θ1 and θ2 form a linear pair, the we can subtract θ2 from 180° to find θ1. So 180° minus 56.7° is 123.7°.

We cannot forget about r. Let's solve for that as well.

42 + 62 = c2

sqrt(42 + 62) = c

sqrt(16 + 36) = c

sqrt(52) = c

sqrt(4) * sqrt(13) = c

2 * sqrt(13) = c

7.21 = c

Pythagorean Theorem

Substituted 4 for a and 6 for b

Squared both numbers

Found square root

Converted to decimal

We have the theta that we want. We also have our r, so now we can use z = r * (cosθ + i * sinθ). Our final answer is z = 2sqrt(13) * (cos127.3° + i * sin127.3°).

### Example 3 Example 3:

Here we have our number in standard form, z = a + b * i. We must convert it to the form z = r * (cosθ + i * sinθ). We know r = sqrt( a2 + b2) and tanθ = b/a.

If we use the a and b of our complex number to form a triangle. The modulus of z or r or the distance between the origin and our point is the hypotenuse. Now we can use a and b as our legs to find our missing side length.

[2sqrt(3)]2 + (-2)2 = c2

sqrt([2sqrt(3)]2 + (-2)2) = c

sqrt(4 * 3 + 4) = c

sqrt(12 + 4) = c

sqrt(16) = c

4 = c

Pythagorean Theorem

Substituted 2sqrt(3) for a and 2 for b

Squared both numbers

Multiplied

Found square root

Now we know c, the hypotenuse, or our r, 4. Now, we must solve for our argument or theta, by using tanθ = b/a. Remember b is the opposite side length while a is the adjacent side length.

tanθ2= b/a

tanθ2 = -2 / [2sqrt(3)]

tanθ2 = -1 / [1sqrt(3)]

tanθ2 = -sqrt(3) / 3

θ2 = tan-1(-sqrt(3) / 3)

θ2 = -π/6

Substituted 2 for a and 4 for b

Divided by 2

Multiplied by sqrt(3)

Use the inverse tangent on both sides

Calculated

Since θ1 and θ2 form a complete circle pair, the we can add 2π and θ2 to find θ1. So 2π plus -π/6 is 11π/6.

We have the theta that we want. We also have our r, so now we can use z = r * (cosθ + i * sinθ). Our final answer is z = 4 * [cos(11π/6) + i * sin(11π/6)].