Question

I'll give brainliest and points to whomever answers these two questions, thank you so much in advance!

Answer 1

Answer:

Q8.  a.  IR

Q9.  b.  -4

Step-by-step explanation:

Q8- what's the S.S of the inequality: x² + 4 > 2

x² + 4 > 2

⇔ x² > 2 - 4

⇔ x² > -2

Then the set of solutions is all real numbers because  x² is always a positive number which means for all real numbers  x² > -2.

……………………………………………………………………………

Q9- if x + yi = (1 + i)⁴ , then

x + yi = [(1 + i)²]² = [2i]² = 2²×i² = 4×(−1) = −4

(Just remember that (1 + i)² = 2i )

Answer 2

Answer:

Answer b of the first one

answer of q9

Step-by-step explanation:

(x+iy)(2−3i)=4+i

2x−(3x)i+(2y)i−3yi

2

=4+i

Real

2x+3y

​

​

+

Imaginary

(2y−3x)

​

​

i=4+i

Comparing the real & imaginary parts,

2x+3y=4--------------------------(1)

2y−3x=1----------------------------(2)

Solving eq(1)  &  eq(2),

4x+6y=8

−9x+6y=3

13x=5⇒x=

13

5

​

y=

13

14

​

∴(x,y)=(

13

5

​

,

13

14

​

)

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SIMILAR QUESTIONS

star-struck

If e

x+iy

=α+iβ, then x+iy is called logarithm of α+iβ to the base e.∴log

e

​

(x+iy)=log

e

​

(re

iθ

) =log

e

​

r+iθ where r is modulus value of x+iy & θ be the argument of x+iy If i

α+iβ)

=α+iβ, then α

2

+β

2

 equals

Hard

View solution

>

The modulus of (1 + i) (1 + 2i) (1 + 3i) is equal to