SAT Complex Numbers Questions & Answers

SAT Complex Numbers Problems and Solutions

Q1. If z = 3 – 4i, then calculate z ² .

Solution:

Given z = 3 – 4i

So, z ² = z.z

= (3 – 4i)(3 – 4i)

= 3(3) – 3(4i) – (4i)(3) + (4i)(4i)

= 9 – 12i – 12i + 16i ² {since i ² = -1}

= 9 – 24i - 16

= -7 – 24i

Therefore, z ² = -7 – 24i.

Q2. If z = (3 – i) ² + [(8 – 5i)/(3 + i)] – 9, express z in the form of x + iy where x and y are real numbers.

Solution:

Given z = (3 – i) ² + [(8 – 5i)/(3 + i)] – 9

= (3) ² – 2(3)(i) + (i ² ) + [(8 – 5i)(3 – i)/ (3 + i)(3 – i)] – 9

= (9 – 6i – 1) + [(24 – 8i – 15i + 5i ² )/ (9 – i2)] – 9

= (8 – 6i) + [(24 – 23i – 5)/(9 + 1)] – 9 {since i ² = -1}

= (8 – 6i) + [(19 – 23i)/10] – 9

= (8 – 6i) + (1.9 – 2.3i) – 9

= 0.8 – 7.9i

This is in the form x + iy where x = 0.8 and y = -7.9.

Q3. Simplify:

Solution:

This simplifies to:

Therefore,

Q4. If z ₁ = 3 + 9i and z ₂ = 2 – i, then find |z ₁ /z ₂ |.

Solution:

Given,

z ₁ = 3 + 9i and z ₂ = 2 – i

z ₁ /z ₂ = (3 + 9i)/(2 – i)

= (3 + 9i)(2 + i)/ (2 – i)(2 + i)

= [6 + 3i + 18i + 9i ² ]/ [4 – i ² ]

= (6 + 21i - 9)/ (4 + 1) {since i ² = -1}

= (-3 + 21i)/5

= -0.6 + 4.2i

Now, |z ₁ /z ₂ | = √[(-0.6) ² + (4.2) ² ]

= √(0.36 + 17.64)

= √18

Q5. If |z ² – 2| = |z ² | + 2, then prove that z lies on an imaginary axis.

Solution:

Let z = x + iy be the complex number.

Now, z ² = z.z = (x + iy)(x + iy)

= x ² + ixy + ixy + (iy) ²

= x ² + 2ixy – y ² {since i ² = -1}

z ² – 2 = x ² + 2ixy – y ² – 2

= (x ² – y ² – 2) + i(2xy)

Thus, |z ² – 2| = √[(x ² – y ² – 2) ² + (2xy) ² ]

= √[(x ² – y ² – 2) ² + 4x ² y ² ]

|z| ² + 2 = [√(x ² + y ² )] ² + 2

= x ² + y ² + 2

Given that,

|z ² – 2| = |z ² | + 2

So, √[(x ² – y ² – 2) ² + 4x ² y ² ] = x ² + y ² + 2

Squaring on both sides, we get;

(x ² – y ² – 2) ² + 4x ² y ² = (x ² + y ² + 2) ²

[x ² – (y ² + 2)] ² + 4x ² y ² = [x ² + (y ² + 2)] ²

[x ² – (y ² + 2)] ² – [x ² + (y ² + 2)] ² + 4x ² y ² = 0

As we know, (a – b) ² – (a + b) ² = -4ab,

-4x ² (y ² + 2) + 4x ² y ² = 0

-4x ² y ² – 8x ² + 4x ² y ² = 0

8x ² = 0

x = 0

Therefore, z lies on the y-axis.

Q6. Find the conjugate of z ₁ – z ₂ if z ₁ = 3 + 4i and z ₂ = 6 + 3i.

Solution:

Given,

z ₁ = 3 + 4i

z ₂ = 6 + 3i

z ₁ – z ₂ = (3 + 4i) – (6 + 3i)

= (3 – 6) + i(4 – 3)

= -3 + i

As we know the conjugate of z = x + iy = x – iy.

Conjugate of z ₁ – z ₂ = -3 – i

Q7. Simplify: i ⁶⁷

Solution:

We know that,

i ² = -1, i ³ = -i, i ⁴ = 1

We can write 67 as: 67 = 4 × 16 + 3

So, i ⁶⁷ = i ^{(4 × 16) + 3}

= i ^{(4 × 16)} . i ³

= 1.i ³

= -i

Therefore, i ⁶⁷ = -i.

Q8. Find real x and y if (x – iy) (4 + 6i) is the conjugate of – 7 – 28i.

Solution:

(x – iy)(4 + 6i) = 4x + 6ix – 4iy – 6yi ²

= 4x + i(6x – 4y) + 6y {since i ² = -1}

= (4x + 6y) + i(6x – 4y)

Given that (x – iy)(4 + 6i) is the conjugate of -7 – 28i.

Here, the conjugate of -7 – 28i = -7 + 28i.

So, 4x + 6y = -7

6x – 4y = 28

Solving these two equations, we get; x = -2 and y = -3.

Q9. Find the relation between a and b if z = a + ib if |(z – 4)/(z + 4)| = 3.

Solution:

Given,

z = a + ib

|(z – 4)/(z + 4)| = 3

|(a + ib – 4)/(a + ib + 4)| = 3

We know,

|z| = √(x ² + y ² )

√[(a – 4) ² + b ² ] = 3(a + 4) + 3ib

Comparing real and imaginary parts,

⇒ √((a – 4) ² + b ² ) = 3a + 12

And 0 = 3b

⇒ b = 0

Substituting the value of b in √((a – 4) ² + b ² ) = 3a + 12, we get;

(a – 4) ² = (3a + 12) ²

a ² – 8a + 16 = 9a ² + 72a + 144

⇒ 8a ² + 80a + 128 = 0

⇒ a ² + 10a + 16 = 0

⇒ (a + 6)(a + 4) = 0

Therefore, a = -6 or a = -4

Q10. If |z + 2| = z + 3 (1 + i), then find z.

Solution:

Let z = x + iy be the complex number.

Given,

|z + 2| = z + 3 (1 + i)

⇒ |x + iy + 2| = x + iy + 3 (1 + i)

We know,

|z| = √(x ² + y ² )

√[(x + 2) ² + y ² ] = (x + 3) + i(y + 3)

Comparing real and imaginary parts,

⇒ √((x + 2) ² + y ² ) = x + 3

And 0 = y + 3

⇒ y = -3

Substituting the value of y in √((x + 2) ² + y ² ) = x + 3, we get;

(x + 2) ² + (-3) ² = (x + 3) ²

x ² + 4x + 4 + 9 = x ² + 6x + 9

⇒ 2x = 4

⇒ x = 2

Therefore, z = x + iy = 2 – 3i.

Practice Problems on SAT Complex Numbers

1. Find the conjugate of the complex number (2 – i)/(2 + i).
2. The complex number z = a + ib, where a and b are real numbers, satisfies the equation z ² + 18 – 35i = 0. Find a and b.
3. Calculate the modulus of the complex number −3√3 – 3i.
4. Simplify: (2 + 7i) + (7 − 3i) − (−8 + 6i)
5. If [(2 – i)/(2 + i)] ¹⁰¹ = a + ib, then find the values of a and b.

Conclusion

Understanding complex numbers is crucial in solving advanced algebra questions in standardized examinations such as the SAT. The problems and solutions provided in this essay cover a systematic way of dealing with the core subject. Practicing such questions will instill confidence in dealing with operations involving complex numbers, modulus, conjugates, and their uses. Such practice will not only strengthen theoretical concepts but also improve problem-solving speed and efficiency. Continue to explore and solve more problems to further cement your math foundation!