Question

I have a test Monday and I HAVE to get complex numbers, I don't want to fail :(

Answer 1

Ignore the sentences, I just need to understand how to solve these questions :(
and each question has to be in answers like Re(z) =
Im(z) =
etc

Answer 2

Wow uhmm i can answer a few question to the hw….. ;)

Answer 3

Can you? I hope it can be correct, and can you also help me step by step too? :DD

Answer 4

ye lemme see wht i cn do .:)

Answer 5

Alrighty thank you~ for 5 they are asking though theres 2 complex numbers (8,-2) and their difference is (-4,-6) so what is their numbers?

Answer 6

uhmm the pic is kinda blurry but what dnt you get about these questions doe

Answer 7

Complex numbers are "binomials" of a sort, and are added, subtracted, and multiplied in a similar way. (Division, which is further down the page, is a bit different.) First, though, you'll probably be asked to demonstrate that you understand the definition of complex numbers.

Solve 3 – 4i = x + yi
Finding the answer to this involves nothing more than knowing that two complex numbers can be equal only if their real and imaginary parts are equal. In other words, 3 = x and –4 = y.

To simplify complex-valued expressions, you combine "like" terms and apply the various other methods you learned for working with polynomials.

Simplify (2 + 3i) + (1 – 6i).
(2 + 3i) + (1 – 6i) = (2 + 1) + (3i – 6i) = 3 + (–3i) = 3 – 3i

Simplify (5 – 2i) – (–4 – i).
(5 – 2i) – (–4 – i)

= (5 – 2i) – 1(–4 – i) = 5 – 2i – 1(–4) – 1(–i)

= 5 – 2i + 4 + i = (5 + 4) + (–2i + i)

= (9) + (–1i) = 9 – i

You may find it helpful to insert the "1" in front of the second set of parentheses (highlighted in red above) so you can better keep track of the "minus" being multiplied through the parentheses.

Simplify (2 – i)(3 + 4i).
(2 – i)(3 + 4i) = (2)(3) + (2)(4i) + (–i)(3) + (–i)(4i)

= 6 + 8i – 3i – 4i2 = 6 + 5i – 4(–1)

= 6 + 5i + 4 = 10 + 5i

For the last example above, FOILing works for this kind of multiplication, if you learned that method. But whatever method you use, remember that multiplying and adding with complexes works just like multiplying and adding polynomials, except that, while x2 is just x2, i2 is –1. You can use the exact same techniques for simplifying complex-number expressions as you do for polynomial expressions, but you can simplify even further with complexes because i2 reduces to the number –1.

thats all the info i cn give :/ hope it helps :)