A student multiplies (4+5i) (3-2i) incorrectly and obtains 12-10i. What is the student's mistake? Select the two correct answers. A. They multiplied the second number by the real part of the first number. B. In the multiplication of the imaginary parts, the student forgot to square the i. C. The student has only multiplied the real parts and the imaginary parts. D. The student has added the real parts and multiplied the imaginary parts. E. The student has multiplied the second number by the imaginary part of the first number.

Step-by-step explanation:

Before we figure out the student mistake, let us find the product of the complex numbers ourselves first.

(4+5i) (3-2i)

open the parenthesis

= 4(3)-4(2i)+3(5i)+5i(-2i)

= 12-8i+15i-10i²

Note that in complex number, i² = -1, hence the expression will become;

= 12-8i+15i-10(-1)

= 12-8i+15i+10

collect like terms by separating the real from imaginary part

= 12+10-8i+15i

= 22+7i

From the students answer i.e 12-10i, it can be concluded that;

In the multiplication of the imaginary parts, the student forgot to square the i and the student has only multiplied the real parts and the imaginary parts 4 and 3 to get 12

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse