Question

If (2 + 3i)2 + (2 − 3i)2 = a + bi, a = and b = .

Answer 1

`(2+3i)^2+(2-3i)^2=(4+12i+9i^2)+(4-12i+9i^2)=8+18i^2=8-18=-10`

so
`a=-10` and
`b=0`.

Answer 2

Answer : The value of a = -10 and b = 0

Step-by-step explanation :

As we are given the expression:

`(2+3i)^2+(2-3i)^2=a+bi`

Now we have to calculate the value of a and b.

`(2+3i)^2+(2-3i)^2=a+bi`

Using identity :

`(a+b)^2=a^2+b^2+2ab`

`(a-b)^2=a^2+b^2-2ab`

`[2^2+(3i)^2+2* 2* i]+[2^2+(3i)^2-2* 2* i]=a+bi`

`[4+9i^2+4i]+[4+9i^2-4i]=a+bi`

`4+9i^2+4i+4+9i^2-4i=a+bi`

`4+9i^2+4+9i^2=a+bi`

`8+18i^2=a+bi`

As we know that,
`i^2=(-1)`

So,

`8+18(-1)=a+bi`

`8-18=a+bi`

`-10=a+bi`

Thus, the value of a = -10 and b = 0