Question $(4+i)(2-5 i)$

Studdy Solution

STEP 1

What is this asking? Multiply these two complex numbers, $(4+i)$ and $(2-5i)$ , and express the result in the standard form $a + bi$ . Watch out! Remember the special property of $i$ : $i^2 = -1$ .
Don't forget to use it!
Also, treat $i$ like a variable when multiplying, just like you would with $x$ .

STEP 2

1. Distribute the terms.
2. Simplify using $i^2 = -1$ .
3. Combine like terms.

STEP 3

Alright, let's FOIL these complex numbers!
Remember, we're treating $i$ just like a variable at this stage.
So, we multiply the first terms, the outer terms, the inner terms, and the last terms.

STEP 4

$(4+i)(2-5i) = (4 \cdot 2) + (4 \cdot (-5i)) + (i \cdot 2) + (i \cdot (-5i))$

STEP 5

Let's simplify each term: $(4 \cdot 2) + (4 \cdot (-5i)) + (i \cdot 2) + (i \cdot (-5i)) = 8 - 20i + 2i - 5i^2$

STEP 6

Now, here's where the magic happens!
Remember that special property of $i$ ?
We know that $i^2 = -1$ .
So, let's substitute $-1$ for $i^2$ in our expression:

STEP 7

$8 - 20i + 2i - 5i^2 = 8 - 20i + 2i - 5(-1)$

STEP 8

$8 - 20i + 2i + 5$

STEP 9

Almost there!
Let's combine the real parts (the numbers without $i$ ) and the imaginary parts (the numbers with $i$ ).

STEP 10

Real parts: $8 + 5 = 13$ Imaginary parts: $-20i + 2i = -18i$

STEP 11

Putting it all together, we get: $13 - 18i$

STEP 12

Our final answer, in the beautiful standard form $a + bi$ , is $13 - 18i$ .

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