Question

(-√5 +2√-4)+(1-√-9)+(2+3i)(2-3i) then find value of a and b​

Answer 1

Answer:

$a=14-\sqrt5$

$b=1$

Step-by-step explanation:

Given:

$(-\sqrt5 +2\sqrt{-4})+(1-\sqrt{-9})+(2+3i)(2-3i)=a+ib$

$(-\sqrt5 +2\sqrt{4(-1)})+(1-\sqrt{9(-1)})+2^2-(3i)^2=a+ib$

$(-\sqrt5 +2\sqrt{2^2i^2})+(1-\sqrt{3^2i^2})+2^2-3^2i^2=a+ib$

$(-\sqrt5 +2(2i))+(1-3i)+4-9(-1)=a+ib$

$(-\sqrt5 +4i)+(1-3i)+4+9=a+ib$

$(-\sqrt5 +4i)+14-3i=a+ib$

$\implies\:(14-\sqrt5)+i=a+ib$

separating real and imaginary parts, we get

$a=14-\sqrt5$

$b=1$

Answer 2

Answer:

Answer:

a=14-\sqrt5a=14−

5

b=1b=1

Step-by-step explanation:

Given:

(-\sqrt5 +2\sqrt{-4})+(1-\sqrt{-9})+(2+3i)(2-3i)=a+ib(−

5

+2

−4

)+(1−

−9

)+(2+3i)(2−3i)=a+ib

(-\sqrt5 +2\sqrt{4(-1)})+(1-\sqrt{9(-1)})+2^2-(3i)^2=a+ib(−

5

+2

4(−1)

)+(1−

9(−1)

)+2

2

−(3i)

2

=a+ib

(-\sqrt5 +2\sqrt{2^2i^2})+(1-\sqrt{3^2i^2})+2^2-3^2i^2=a+ib(−

5

+2

2

2

i

2

)+(1−

3

2

i

2

)+2

2

−3

2

i

2

=a+ib

(-\sqrt5 +2(2i))+(1-3i)+4-9(-1)=a+ib(−

5

+2(2i))+(1−3i)+4−9(−1)=a+ib

(-\sqrt5 +4i)+(1-3i)+4+9=a+ib(−

5

+4i)+(1−3i)+4+9=a+ib

(-\sqrt5 +4i)+14-3i=a+ib(−

5

+4i)+14−3i=a+ib

\implies\:(14-\sqrt5)+i=a+ib⟹(14−

5

)+i=a+ib

separating real and imaginary parts, we get

a=14-\sqrt5a=14−

5

b=1b=1