Question

1. find the difference between $(-17 - i)$ and $(2 + 9i)$
2. find the sum of $(26 + 8i)$, $(-13 - 21i)$, and $(-10 + 14i)$
3. find the product of $(15 + i)$ and $(-3 - 7i)$

simplify the following expressions

1. $(13 - 5i) + (-6 + 21i) - (-15 - 11i)$ \t\t5) $(6 + i)(-8 - i)$
2. $(17 - 7i)^2$ \t\t\t\t\t\t7) $(4 - 18i)(4 + 18i)$
3. $3(1 - 2i) + 5i$ \t\t\t\t\t9) $7i + 4(3 - i)$

Explanation:

Problem 1: Find the difference between \((-17 - i)\) and \((2 + 9i)\)

Step 1: Recall the formula for the difference of complex numbers \((a + bi)-(c + di)=(a - c)+(b - d)i\)

Here, \(a=-17\), \(b = - 1\), \(c = 2\), \(d=9\)

Step 2: Substitute the values into the formula

\((-17 - i)-(2 + 9i)=(-17-2)+(-1 - 9)i\)

Step 3: Simplify the real and imaginary parts

\(-17-2=-19\) and \(-1 - 9=-10\)
So, \((-17 - i)-(2 + 9i)=-19-10i\)

Step 1: Recall the formula for the sum of complex numbers \((a + bi)+(c + di)+(e+fi)=(a + c+e)+(b + d + f)i\)

Here, \(a = 26\), \(b = 8\), \(c=-13\), \(d=-21\), \(e=-10\), \(f = 14\)

Step 2: Substitute the values into the formula

\((26 + 8i)+(-13-21i)+(-10 + 14i)=(26-13-10)+(8-21 + 14)i\)

Step 3: Simplify the real and imaginary parts

\(26-13-10 = 3\) and \(8-21 + 14=1\)
So, \((26 + 8i)+(-13-21i)+(-10 + 14i)=3 + i\)

Step 1: Use the distributive property (FOIL method) \((a + bi)(c + di)=ac+adi+bci+bdi^{2}\), and recall that \(i^{2}=-1\)

\((15 + i)(-3-7i)=15\times(-3)+15\times(-7i)+i\times(-3)+i\times(-7i)\)

Step 2: Simplify each term

\(15\times(-3)=-45\), \(15\times(-7i)=-105i\), \(i\times(-3)=-3i\), \(i\times(-7i)=-7i^{2}=7\) (since \(i^{2}=-1\))

Step 3: Combine like terms

Real parts: \(-45 + 7=-38\)
Imaginary parts: \(-105i-3i=-108i\)
So, \((15 + i)(-3-7i)=-38-108i\)

Answer:

\(-19 - 10i\)

Problem 2: Find the sum of \((26 + 8i)\), \((-13-21i)\), and \((-10 + 14i)\)