1.3 Given: $$ (a+b i)(3-4 i)=-6-17 i $$ Solve $$ a+b i $$ Given: $$ 5 x^{2} i-4=3 y i^{4}+y^{2} i^{2}-y(x i+i) $$ Solve $$ x $$ and $$ y $$ where $$ y \in N $$.

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UpStudy AI · Accepted Answer

We start with the two given equations.

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**Problem 1**

We are given
$$
(a+bi)(3-4i) = -6 - 17i.
$$
Our goal is to find $$a$$ and $$b$$.

1. Multiply out the left-hand side:
   $$
   (a+bi)(3-4i) = a(3-4i) + bi(3-4i) = 3a - 4ai + 3bi - 4bi^2.
   $$
2. Recall that $$i^2 = -1$$, so:
   $$
   -4bi^2 = 4b.
   $$
   Thus, the expression becomes:
   $$
   (3a+4b) + (-4a+3b)i.
   $$
3. Set the real and imaginary parts equal to the corresponding parts on the right-hand side:
   $$
   \begin{cases}
   3a + 4b = -6, \
   -4a + 3b = -17.
   \end{cases}
   $$
4. Solve the system:
   - Multiply the first equation by 3:
     $$
     9a + 12b = -18.
     $$
   - Multiply the second equation by 4:
     $$
     -16a + 12b = -68.
     $$
   - Subtract the first new equation from the second:
     $$
     (-16a+12b) - (9a+12b) = -68 - (-18) \quad \Longrightarrow \quad -25a = -50,
     $$
     so
     $$
     a = 2.
     $$
   - Substitute $$a=2$$ into $$3a+4b=-6$$:
     $$
     3(2) + 4b = -6 \quad \Longrightarrow \quad 6 + 4b = -6 \quad \Longrightarrow \quad 4b = -12,
     $$
     hence,
     $$
     b = -3.
     $$
5. Therefore, the solution is:
   $$
   a+bi=2-3i.
   $$

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**Problem 2**

We are given
$$
5x^{2}i - 4 = 3y\,i^{4} + y^{2}\,i^{2} - y\,(xi + i),
$$
and we need to solve for $$x$$ and $$y$$ with $$y \in \mathbb{N}$$.

1. First, simplify the powers of $$i$$:
   - Since $$i^4=1$$,
     $$
     3y\, i^{4} = 3y.
     $$
   - And $$i^2=-1$$,
     $$
     y^{2}\, i^{2} = - y^{2}.
     $$
   Thus, the equation becomes:
   $$
   5x^2 i - 4 = 3y - y^2 - y\,(xi + i).
   $$
2. Factor the last term:
   $$
   -y\,(xi + i) = -y\,i(x+1).
   $$
   Now, the equation is:
   $$
   5x^2 i - 4 = (3y - y^2) - y\,i(x+1).
   $$
3. Separate the equation into its real and imaginary parts.

- The **real part** on the left is $$-4$$ and on the right is $$3y - y^2$$:
     $$
     3y - y^2 = -4.
     $$
   - The **imaginary part** on the left is $$5x^2$$ (the coefficient of $$i$$) and on the right is $$-y(x+1)$$:
     $$
     5x^2 = - y(x+1).
     $$

4. Solve the real part:
   $$
   3y - y^2 = -4.
   $$
   Rearranging,
   $$
   y^2 - 3y - 4 = 0.
   $$
   Factor or use the quadratic formula:
   $$
   y = \frac{3 \pm \sqrt{9+16}}{2} = \frac{3 \pm 5}{2}.
   $$
   So, the solutions are:
   $$
   y = 4 \quad \text{or} \quad y = -1.
   $$
   Since $$y$$ must be a natural number, we take:
   $$
   y=4.
   $$

5. Now substitute $$y=4$$ into the imaginary part:
   $$
   5x^2 = -4(x+1).
   $$
   Rearranging,
   $$
   5x^2 + 4x + 4 = 0.
   $$
6. Solve the quadratic equation for $$x$$ using the quadratic formula:
   $$
   x = \frac{-4 \pm \sqrt{(4)^2 - 4\cdot 5\cdot 4}}{2\cdot 5} = \frac{-4 \pm \sqrt{16 - 80}}{10} = \frac{-4 \pm \sqrt{-64}}{10}.
   $$
   Since $$\sqrt{-64}=8i$$, we have:
   $$
   x = \frac{-4 \pm 8i}{10} = \frac{-2 \pm 4i}{5}.
   $$

Thus, the solutions are:
$$
y=4 \quad \text{and} \quad x=\frac{-2+4i}{5} \quad \text{or} \quad x=\frac{-2-4i}{5}.
$$

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Final Answers:

$$
a+bi=2-3i,
$$
$$
y=4 \quad \text{and} \quad x=\frac{-2 \pm 4i}{5}.
$$
$$ a = 2 $$ and $$ b = -3 $$, so $$ a + bi = 2 - 3i $$. For the second problem, $$ y = 4 $$ and $$ x = \frac{-2 \pm 4i}{5} $$.

1.3 Given: \( (a+b i)(3-4 i)=-6-17 i \) Solve \( a+b i \) Given: \( 5 x^{2} i-4=3 y i^{4}+y^{2} i^{2}-y(x i+i) \) Solve \( x \) and \( y \) where \( y \in N \).

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