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Math  /  Algebra

Question111(3+i)(5−i)111(3+i)(5-i)

Studdy Solution

STEP 1

What is this asking? We're multiplying a real number by two complex numbers! Watch out! Remember the distributive property and how to multiply complex numbers correctly, especially how to handle i2i^2!

STEP 2

1. Multiply the complex numbers
2. Multiply by the real number

STEP 3

Alright, let's **multiply** those complex numbers (3+i)(3+i) and (5−i)(5-i) first.
This will make our lives much easier later on.

STEP 4

Remember how to **distribute**, or FOIL, to multiply these binomials?
It's like shaking hands with everyone at a party!
We multiply each term in the first parentheses by each term in the second parentheses.
So, we get: (3+i)(5−i)=3⋅5+3⋅(−i)+i⋅5+i⋅(−i)(3+i)(5-i) = 3 \cdot 5 + 3 \cdot (-i) + i \cdot 5 + i \cdot (-i)

STEP 5

Let's **simplify** this a bit: 15−3i+5i−i215 - 3i + 5i - i^2

STEP 6

Now, **combine** those *i* terms: 15+2i−i215 + 2i - i^2

STEP 7

Remember the magical property of ii? i2i^2 is equal to **-1**!
Let's **substitute** that in: 15+2i−(−1)15 + 2i - (-1)

STEP 8

This **simplifies** to: 15+2i+115 + 2i + 1

STEP 9

And finally, **combining** the constants gives us: 16+2i16 + 2i

STEP 10

Great! Now we just need to **multiply** our result 16+2i16 + 2i by **111**.

STEP 11

Let's **distribute** that 111: 111(16+2i)=111⋅16+111⋅2i111(16 + 2i) = 111 \cdot 16 + 111 \cdot 2i

STEP 12

Time to **multiply**: 1776+222i1776 + 222i

STEP 13

Our **final answer** is 1776+222i1776 + 222i!

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