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If z = -5 + 3i, find the value of (z^(4)...

If `z = -5 + 3i`, find the value of `(z^(4)+9z^(3)+26z^(2)-14z + 8)`.

A

`60`

B

`-60`

C

`-48`

D

`48`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the expression \( z^4 + 9z^3 + 26z^2 - 14z + 8 \) where \( z = -5 + 3i \), we will calculate each power of \( z \) step by step. ### Step 1: Calculate \( z^2 \) \[ z^2 = (-5 + 3i)^2 = (-5)^2 + 2(-5)(3i) + (3i)^2 = 25 - 30i - 9 = 16 - 30i \] **Hint:** Use the formula \( (a + bi)^2 = a^2 + 2abi + (bi)^2 \) to expand the square. ### Step 2: Calculate \( z^3 \) \[ z^3 = z \cdot z^2 = (-5 + 3i)(16 - 30i) \] \[ = -5 \cdot 16 + (-5)(-30i) + 3i \cdot 16 + 3i \cdot (-30i) \] \[ = -80 + 150i + 48i + 90 = 10 + 198i \] **Hint:** Multiply the complex numbers using the distributive property and remember that \( i^2 = -1 \). ### Step 3: Calculate \( z^4 \) \[ z^4 = z \cdot z^3 = (-5 + 3i)(10 + 198i) \] \[ = -5 \cdot 10 + (-5)(198i) + 3i \cdot 10 + 3i \cdot 198i \] \[ = -50 - 990i + 30i + 594 = 544 - 960i \] **Hint:** Again, use the distributive property to multiply the complex numbers. ### Step 4: Substitute into the expression Now we substitute \( z^4 \), \( z^3 \), \( z^2 \), and \( z \) into the expression \( z^4 + 9z^3 + 26z^2 - 14z + 8 \): \[ = (544 - 960i) + 9(10 + 198i) + 26(16 - 30i) - 14(-5 + 3i) + 8 \] Calculating each term: 1. \( 9z^3 = 9(10 + 198i) = 90 + 1782i \) 2. \( 26z^2 = 26(16 - 30i) = 416 - 780i \) 3. \( -14z = -14(-5 + 3i) = 70 - 42i \) Now combine all the terms: \[ = (544 + 90 + 416 + 70 + 8) + (-960i + 1782i - 780i - 42i) \] Calculating the real part: \[ 544 + 90 + 416 + 70 + 8 = 1128 \] Calculating the imaginary part: \[ -960 + 1782 - 780 - 42 = 0 \] ### Final Result Thus, the value of \( z^4 + 9z^3 + 26z^2 - 14z + 8 \) is: \[ \boxed{1128} \]

To solve the expression \( z^4 + 9z^3 + 26z^2 - 14z + 8 \) where \( z = -5 + 3i \), we will calculate each power of \( z \) step by step. ### Step 1: Calculate \( z^2 \) \[ z^2 = (-5 + 3i)^2 = (-5)^2 + 2(-5)(3i) + (3i)^2 = 25 - 30i - 9 = 16 - 30i \] ...
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