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MAT1511 Assignment 2 (Answers) Semester 1 - 2023 R50,00 Add to cart

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MAT1511 Assignment 2 (Answers) Semester 1 - 2023

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Course
MAT1511 - Precalculus Mathematics B (MAT1511)

Institution
University Of South Africa (Unisa)

MAT1511 Assignment 2 (Answers) Semester 1 - 2023 Questions asked: 1. Write (2+3i) (4−3i) (3+2i) 2 (4) in the form a+bi, where a,b ∈ R. 2. Given z1 = 2∠180o , ; z2 = 3∠270o and z3 = 1∠180o . Determine the following and leave your answers in rectangular form: (i) (z1) 2 +...

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where a
mat1511 assignment 2 answers semester 1 2023
1 write 23i 4−3i 32i 2 4 in the form abi
b ∈ r 2 given z1 2∠180o
z2 3∠270o and z3 1∠180o determine the following an

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By: dimakatsomosadiapelo • 1 year ago

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, 1. Write (2+3i) (4−3i) (3+2i) 2 (4) in the form a+bi, where a,b ∈ R.
To simplify the expression, let's perform the calculations step by step:

1. (2 + 3i) * (4 - 3i):

Using the FOIL method (First, Outer, Inner, Last), we can expand the expression as follows:

= 2 * 4 + 2 * (-3i) + 3i * 4 + 3i * (-3i)
= 8 - 6i + 12i - 9i^2

Remember that i^2 is equal to -1. Substituting it into the equation:

= 8 - 6i + 12i - 9(-1)
= 8 - 6i + 12i + 9
= 17 + 6i

2. (3 + 2i) / (4):

Dividing the complex number by a real number is straightforward. We simply divide each term by 4:

= (3/4) + (2i/4)
= 3/4 + i/2

Now, let's proceed to the next step:

3. (17 + 6i) * (3/4 + i/2):

Again, using the FOIL method, we can expand the expression:

= (17 * 3/4) + (17 * i/2) + (6i * 3/4) + (6i * i/2)
= 51/4 + 17i/2 + 18i/4 - 6
= 51/4 + 17i/2 + 9i/2 - 6
= (51 + 17i + 9i - 24) / 4
= (27 + 26i) / 4

Finally, we divide each term by 4:

= 27/4 + 26i/4
= 27/4 + 13/2i

Therefore, the expression (2+3i)(4-3i) / (3+2i)(4) can be simplified to 27/4 + 13/2i.

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