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

Answer: (B)

$75$

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