Answer : The value of a = -10 and b = 0
Step-by-step explanation :
As we are given the expression:
[tex](2+3i)^2+(2-3i)^2=a+bi[/tex]
Now we have to calculate the value of a and b.
[tex](2+3i)^2+(2-3i)^2=a+bi[/tex]
Using identity :
[tex](a+b)^2=a^2+b^2+2ab[/tex]
[tex](a-b)^2=a^2+b^2-2ab[/tex]
[tex][2^2+(3i)^2+2\times 2\times i]+[2^2+(3i)^2-2\times 2\times i]=a+bi[/tex]
[tex][4+9i^2+4i]+[4+9i^2-4i]=a+bi[/tex]
[tex]4+9i^2+4i+4+9i^2-4i=a+bi[/tex]
[tex]4+9i^2+4+9i^2=a+bi[/tex]
[tex]8+18i^2=a+bi[/tex]
As we know that, [tex]i^2=(-1)[/tex]
So,
[tex]8+18(-1)=a+bi[/tex]
[tex]8-18=a+bi[/tex]
[tex]-10=a+bi[/tex]
Thus, the value of a = -10 and b = 0
