Question

(7+2i)^2 (6+8i)*(6-8i)

Answer 1

(7+2i)^2 =9²=81

(6+8i)*(6-8i)=14*(-2)=-28

или по формуле разность квадратоа 6²-8²=36-64=-28

Answer 2

Let's solve the given expression step by step:

1. (7+2i)^2:

To square a complex number (a + bi)^2, you can use the FOIL method (First, Outer, Inner, Last). It is done as follows:

(a + bi)^2 = a^2 + 2abi + (bi)^2

where (bi)^2 is equal to -b^2.

So, for (7 + 2i)^2:

(7 + 2i)^2 = 7^2 + 2 * 7 * 2i + (2i)^2 = 49 + 28i - 4 = 45 + 28i

2. (6+8i)*(6-8i):

To multiply these complex numbers, you can also use the FOIL method:

(6 + 8i)*(6 - 8i) = 6 * 6 + 6 * (-8i) + 8i * 6 + 8i * (-8i)

Remember that (8i)*(8i) is equal to -64.

So, for (6 + 8i)*(6 - 8i):

(6 + 8i)*(6 - 8i) = 36 - 48i + 48i - 64 = 36 - 64 = -28

Now, let's multiply the two results we obtained in step 1 and step 2:

(45 + 28i) * (-28)

To calculate this, simply distribute the (-28) to each term:

= 45*(-28) + 28i*(-28)

= -1260 - 784i

So the final result is -1260 - 784i.