Question

help for jee mains anz​

Answer 1

Answer:

I think options are wrong. Correct answer is 7/10 + (-9/10)

Answer 2

Answer:

Third Option

Step-by-step explanation:

Given :-

(2+i)²/(3+i)

To find :-

Find the conjugate of (2+i)²/(3+i) in the form of a+i b ?

Solution :-

Given that

(2+i)²/(3+i)

On multiplying both numerator and the denominator with (3-i) then

=> [(2+i)²/(3+i)]×[(3-i)/(3-i)]

=> [(2+i)²(3-i)]/[(3+i)(3-i)]

=> [(2+i)²(3-i)]/[(3²-i²)

Since (a+b)(a-b) = a²-b²

Where, a = 3 and b = i

=> [(2+i)²(3-i)]/(9-i²)

=> [(2+i)²(3-i)]/[9-(-1)]

Since ,i² = -1

=> [(2+i)²(3-i)]/(9+1)

=> [(2+i)²(3-i)]/10

=> [{2²+i²+2(2)(i)}(3-i)]/10

Since (a+b)² = a²+2ab+b²

Where, a = 2 and b = i

=> (4+i²+4i)(3-i)/10

=> (4-1+4i)(3-i)/10

=> (3+4i)(3-i)/10

=> (9-3i+12i-4i²)/10

=> (9+9i-4(-1))/10

Since, i² = -1

=> (9+9i+4)/10

=> (13+9i)/10

=> (13/10)+ i (9/10)

We know that

The conjugate of a+ib is a-ib

So,

The conjugate of (13/10)+ i (9/10) is

(13/10)-i(9/10) => (13/10)+i (-9/10)

Answer:-

The conjugate of (13/10)+ i (9/10) is (13/10)+i (-9/10)

Used formulae:-

→The conjugate of a+ib is a-ib

→ (a+b)(a-b) = a²-b²

→ (a+b)² = a²+2ab+b²

→ i² = -1