Multiplying complex numbers?

Question

[ Putting in a=bi form]
I forgot to add
1. (2-i)(4+2i)
i dont know why but i keep screwing up this problem. someone please explain it. Much love<33 thankyou.

Answer 1

Simplify the expression & place the final solution in the a+bi form.

(2 - i)(4 + 2i)

First: eliminate the parenthesis by using the Foil Method.

(2)(4)+ (2)(2i)+ (-i)(4)+ (-i)(2i)

8 + 4i+ (- 4i)+ (- 2i^2)

8 + 4i - 4i - 2i^2

Sec: combine "like" terms.

8 + 0i - 2i^2

8 - 2i^2

Third: the i^2 becomes -1 > replace -1 with i^2

8 - 2(-1)

8 + 2

10

Answer 2

Remember that i^2 = -1 in complex numbers.

(2 - i)(4 + 2i) = 8 - 4i +4i - 2i^2 obtained just multiplying through.

(2 - i)(4 + 2i) = 8 - 2(-1) = 8+ 2 = 10

Answer 3

Just multiply it out using FOIL, or the distributive property, or whatever you want to call it.

(2 - i)(4 + 2i) = 8 - 4i + 4i - 2i^2 = 8 - 2i^2

Now, since i^2 = -1, we can further simplify:

8 - 2i^2 = 8 - 2(-1) = 8 + 2 = 10.

So the answer is 10. (or 10 + 0i, if you like).

Answer 4

So this is just like multiplying expressions - use FOIL

8 -4i +4i - 2i^2 =

8 - 2i^2

i^2 = -1, by definition, so

8 + 2

10, or 10 +0i

Answer 5

(2 -i ) (4 +2i)

= 8 -4i + 4i -2i^2

= 8 + 0 -2(-1)

= 8 + 2

= 10


Nota : i^2 = -1

Answer 6

You must distribute.
2*4 + 2*2i - 4i -2i^2
8 + 4i - 4i - (2*-1) , because (i^2 = -1)
8 - -2 = 10

Answer 7

(2-i)(4+2i)
= 2(4+2i)-i(4+2i)
= 2(4)+2(2i) -i(4)-i(2i)
= 8 + 4i -4i +1(2)
= 8+2
= 10

Answer 8

Again, use FOIL: You get
8 -2i² -4i + 4i= 8+2 = 10.

Answer 9

(2-i)(4+2i): multiply it out just like in algebra
= 2(4) + (-i)(4) + 2(2i) + (-i)(2i)
= 8 + 0i - 2i^2
= 8 + 2
= 10.

Answer 10

For the first one you've been close, the answer is 14-18i. I were given this by technique of creating use of the FOIL technique, (4+2i)(a million-5i) = 4-20i+2i-10i^2, upload all like words to get 14-18i, out of your answer is sounds like you in elementary words added incorrect. For the 2d one the answer I were given replaced into (9-12i)/5. To get this take the equation (6-3i)/(2+i) and mult. by technique of the complicated conjugate of the denominator so.. (6-3i)/(2+i) * (2-i)/(2-i) by technique of doing this it really is going to eliminate the i words contained in the denominator and provide you with a sparkling complicated huge style contained in the numerator. by technique of doing this I were given (9-12i)/5