Question

Let \( z_1 \) and \( z_2 \) be two complex numbers such that \( z_1 + z_2 = 5 \) and \( z_1^3 + z_2^3 = 20 + 15i \). Then \( \left| z_1^4 + z_2^4 \right| \) equals

• A. \( 30 \sqrt{3} \)
• B. \( 75 \)
• C. \( 15 \sqrt{15} \)
• D. \( 25 \sqrt{3} \)

Answer 1

The Correct Option is B

To solve this problem, we need to determine the magnitude of \( |z_1^4 + z_2^4| \) given the conditions \( z_1 + z_2 = 5 \) and \( z_1^3 + z_2^3 = 20 + 15i \).

First, recall the identity for the sum of cubes:

\(z_1^3 + z_2^3 = (z_1 + z_2)(z_1^2 - z_1z_2 + z_2^2)\)

Substituting \(z_1 + z_2 = 5\), we get:

\(20 + 15i = 5(z_1^2 - z_1z_2 + z_2^2)\)

Therefore,

\(z_1^2 - z_1z_2 + z_2^2 = 4 + 3i\)

Also, from the identity \((z_1 + z_2)^2 = z_1^2 + 2z_1z_2 + z_2^2\), we have:

\(25 = z_1^2 + 2z_1z_2 + z_2^2\)

Let's subtract these results:

\(25 - (4 + 3i) = z_1^2 + 2z_1z_2 + z_2^2 - (z_1^2 - z_1z_2 + z_2^2)\)

This simplifies to:

\(21 - 3i = 3z_1z_2\)

Thus,

\(z_1z_2 = 7 - i\)

Now, let's use the identity for the sum of fourth powers:

\(z_1^4 + z_2^4 = (z_1^2 + z_2^2)^2 - 2(z_1z_2)^2\)

We already know:

\(z_1^2 + z_2^2 = 4 + 3i + z_1z_2 = 4 + 3i + (7 - i) = 11 + 2i\)

Now compute \((z_1^2 + z_2^2)^2\):

\((11 + 2i)^2 = 121 + 44i + 4i^2 = 121 + 44i -4 = 117 + 44i\)

Next, compute \(2(z_1z_2)^2\):

\((z_1z_2)^2 = (7 - i)^2 = 49 - 14i + i^2 = 49 - 14i - 1 = 48 - 14i\)

Therefore,

\(2(z_1z_2)^2 = 2(48 - 14i) = 96 - 28i\)

Substitute back,

\(z_1^4 + z_2^4 = (117 + 44i) - (96 - 28i) = 21 + 72i\)

We need the magnitude:

\(\left|z_1^4 + z_2^4\right| = \sqrt{21^2 + 72^2} = \sqrt{441 + 5184} = \sqrt{5625} = 75\)

Thus, the answer is \( 75 \).

Answer 2

Given:

\[ z_1 + z_2 = 5 \quad \text{and} \quad z_1^3 + z_2^3 = 20 + 15i \]

Let \( S = z_1 + z_2 \) and \( P = z_1 z_2 \). We know that:

\[ S = 5 \]

Using the identity for the sum of cubes:

\[ z_1^3 + z_2^3 = (z_1 + z_2)\left(z_1^2 - z_1 z_2 + z_2^2\right) \]

Since \( z_1^2 + z_2^2 = S^2 - 2P \), we can write:

\[ z_1^3 + z_2^3 = S(S^2 - 3P) = 20 + 15i \]

Substitute \( S = 5 \):

\[ 5(25 - 3P) = 20 + 15i \]

Solving for \( P \), we get:

\[ 125 - 15P = 20 + 15i \] \[ 15P = 105 - 15i \] \[ P = 7 - i \]

Now we need to find \( z_1^4 + z_2^4 \). Using the identity:

\[ z_1^4 + z_2^4 = (z_1^2 + z_2^2)^2 - 2(z_1 z_2)^2 \]

Since \( z_1^2 + z_2^2 = S^2 - 2P \), we have:

\[ z_1^2 + z_2^2 = 5^2 - 2(7 - i) = 25 - 14 + 2i = 11 + 2i \]

Now, square \( z_1^2 + z_2^2 \):

\[ (z_1^2 + z_2^2)^2 = (11 + 2i)^2 = 121 + 44i + 4i^2 = 121 + 44i - 4 = 117 + 44i \]

Next, calculate \( (z_1 z_2)^2 \):

\[ (z_1 z_2)^2 = (7 - i)^2 = 49 - 14i + i^2 = 49 - 14i - 1 = 48 - 14i \]

Thus,

\[ z_1^4 + z_2^4 = (117 + 44i) - 2(48 - 14i) \] \[ = 117 + 44i - 96 + 28i \] \[ = 21 + 72i \]

Finally, we find \( |z_1^4 + z_2^4| \):

\[ |z_1^4 + z_2^4| = \sqrt{21^2 + 72^2} = \sqrt{441 + 5184} = \sqrt{5625} = 75 \]

The answer is: 75