Question

Q56. If
(4 + 3i)(1 + 4i) = x + iy. Then (x,y) =
(A) (-8,-19)
(B) (-8,19)
(C) (8,-19)
(D) (8,19​

Answer 1

Answer :

(C) (8 , -19)

Solution :

• Given : (4 + 3i)(1 + 4i) = x + iy
• To find : (x , y) = ?

We have ;

=> (4 + 3i)(1 + 4i) = x + iy

=> 4•1 + 4•4i + 3i•1 + 3i•4i = x + iy

=> 4 + 16i + 3i + 12i² = x + iy

=> 4 + 19i + 12•(-1) = x + iy

=> 4 + 19i - 12 = x + iy

=> -8 + 19i = x + yi

Now ,

Comparing both sides , we get ;

x = -8

y = 19

Hence ,

(x , y) = (-8 , 19)

Answer 2

⛥ $\rm{\: Hey \: Mate \: !}$

This is the first time I am dealing with $\bold{COMPLEX \: \: NUMBERS}$. So I am confirm that this question will be really memorable for me. Thanks a lot for asking this question. :)

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\large{\underline{\underline{\sf{\maltese\:\: \red{Question \: : } }}}}$

If (4 + 3i)(1 + 4i) = x + iy. Then (x,y) =

(A) (-8,-19)

(B) (-8,19)

(C) (8,-19)

(D) (8,19​)

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\large{\underline{\underline{\sf{\maltese\:\: \blue{Concept \: Used \: : } }}}}$

√-1 is called iota. It is shortened as $\bold{i}$.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\large{\underline{\underline{\sf{\maltese\:\: \green{Solution \: : } }}}}$

(4 + 3i)(1 + 4i) = x + iy

➫ 4(1 + 4i) + 3i(1 + 4i) = x + iy

➫ 4 + 16i + 3i - 12 = x + iy

➫ -8 + 19i = x + iy

Now comparing both sides we get,

★ -8 = x

→ x = - 8

★ 19i = iy

→ 19 = y

→ y = 19

So, (x , y) = (-8 , 19)

So, $\bold{Option \: (B) }$ is correct.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\large{\underline{\underline{\sf{\maltese\:\: \orange{Answer \: : } }}}}$

$\bold{Option \: (B) }$ is correct.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬