Complex Number Questions

In summary, complex numbers are numbers with both a real and imaginary part written in the form a + bi. They are used in various fields of science and have different applications compared to real numbers. To add and subtract complex numbers, you add and subtract their real and imaginary parts separately. To multiply and divide complex numbers, you can use the FOIL method and the conjugate of the denominator, respectively.
  • #1
teetar
35
0
Please, do not give me answers to these questions, I just want to have some hints/tips to point me in the correct direction, or if you're so inclined, you can use another similar/relevant to help me understand this concept. Thanks!

Homework Statement


a.) [/B]Given that z/(z + 2) = 2-i, z ∈ ℂ, find z in the form a+bi
b.) Given that (a + bi)2 = 3 + 4i obtain a pair of simultaneous equations involving a and b. Hence find the two square roots of 3 + 4i.

Homework Equations


None, should be basic logic/algebra.

The Attempt at a Solution


a.)[/B]
z = (z + 2)(2 - i) = 2z - zi + 4 - 2i = 2z + 4 - zi - 2i
Is it then right to say that:
z = a + bi ∴ a = 2z + 4, b = -z - 2 ?
However, that's not very helpful.
I could plug z into the original equation and try to do some long algebra:
2 - i = z/(z + 2) = (2z + 4 - zi - 2i)/(2z + 4 - zi - 2i + 2)
2 - i = (2z + 4 - zi - 2i) / (2z + 6 - zi - 2i)
2 - i = [2z + 4 - (z + 2)i] / [2z + 6 - (z - 2)i]
I could at this point multiply by the conjugate to get reals out of the denominator but that's a lot of work and I feel like I've already missed the point of this, as the question isn't worth too many points.
So, the most meaningful information I've presented relevant to this question (I think) is:
z = 2z + 4 - zi - 2i
2 - i = [2z + 4 - (z + 2)i] / [2z + 6 - (z - 2)i]

b.)
(a + bi)2 = (a + bi)(a + bi) = a2 + 2abi + b2
a2 + 2abi + b2 = 3 + 4i
a2 + b2 + 2abi = 3 + 4i
Hence, is it right to say that:
a2 + b2 = 3, 2ab = 4, so ab = 2 ?
This would mean that a = 2/b and likewise, b = 2/a
I can try substituting this into the other equation:
(2/b)2 + b2 = 3
(4/b2) + b2 = 3
4 + b4 = 3b2
4 = 3b2 - b4
Unfortunately, from here I'm completely lost, even though it's basic algebra, I can't quite put my finger on the next move. In fact, I'm not even sure if I've been doing anything remotely correct to this point.

Thank you all for any help you can offer me!
 
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  • #2
teetar said:
Please, do not give me answers to these questions, I just want to have some hints/tips to point me in the correct direction, or if you're so inclined, you can use another similar/relevant to help me understand this concept. Thanks!

Homework Statement


a.) [/B]Given that z/(z + 2) = 2-i, z ∈ ℂ, find z in the form a+bi
b.) Given that (a + bi)2 = 3 + 4i obtain a pair of simultaneous equations involving a and b. Hence find the two square roots of 3 + 4i.

Homework Equations


None, should be basic logic/algebra.

The Attempt at a Solution


a.)[/B]
z = (z + 2)(2 - i) = 2z - zi + 4 - 2i = 2z + 4 - zi - 2i
Is it then right to say that:
z = a + bi ∴ a = 2z + 4, b = -z - 2 ?
No. You have z = 2z -zi + 4 - 2i
Move all the z terms to one side, and solve algebraically for z.
teetar said:
However, that's not very helpful.
I could plug z into the original equation and try to do some long algebra:
2 - i = z/(z + 2) = (2z + 4 - zi - 2i)/(2z + 4 - zi - 2i + 2)
2 - i = (2z + 4 - zi - 2i) / (2z + 6 - zi - 2i)
2 - i = [2z + 4 - (z + 2)i] / [2z + 6 - (z - 2)i]
I could at this point multiply by the conjugate to get reals out of the denominator but that's a lot of work and I feel like I've already missed the point of this, as the question isn't worth too many points.
So, the most meaningful information I've presented relevant to this question (I think) is:
z = 2z + 4 - zi - 2i
2 - i = [2z + 4 - (z + 2)i] / [2z + 6 - (z - 2)i]

b.)
(a + bi)2 = (a + bi)(a + bi) = a2 + 2abi + b2
You have a mistake right off the bat.
(bi)(bi) ##\neq## b2
teetar said:
a2 + 2abi + b2 = 3 + 4i
a2 + b2 + 2abi = 3 + 4i
Hence, is it right to say that:
a2 + b2 = 3, 2ab = 4, so ab = 2 ?
This would mean that a = 2/b and likewise, b = 2/a
I can try substituting this into the other equation:
(2/b)2 + b2 = 3
(4/b2) + b2 = 3
4 + b4 = 3b2
4 = 3b2 - b4
Unfortunately, from here I'm completely lost, even though it's basic algebra, I can't quite put my finger on the next move. In fact, I'm not even sure if I've been doing anything remotely correct to this point.

Thank you all for any help you can offer me!
 
  • #3
For part b, you were on the right track by equating the real parts and imaginary parts. That is, assuming a and b are purely real.
Fix the sign error pointed out above and you should be in the clear.
If you again come to a point where you have a problem like ##b^4+Cb^2+D=0##, remember to substitute ##b^2=x## and solve like any other quadratic.
 

1. What are complex numbers?

Complex numbers are numbers that consist of a real part and an imaginary part. They are written in the form a + bi, where a is the real part and bi is the imaginary part with i representing the square root of -1.

2. How are complex numbers used in science?

Complex numbers are used in various fields of science, such as physics, engineering, and mathematics. They are particularly useful in solving problems involving alternating currents, signal processing, quantum mechanics, and many more.

3. What is the difference between a real number and a complex number?

A real number is a number that can be represented on a number line, while a complex number has both a real and an imaginary part. Real numbers are used to represent quantities like distance, time, and temperature, while complex numbers are used to represent quantities that involve both magnitude and direction, such as electrical current and force.

4. How do you add and subtract complex numbers?

To add complex numbers, you simply add their real parts and imaginary parts separately. For example, (3 + 2i) + (5 + 4i) = (3+5) + (2i+4i) = 8 + 6i. To subtract complex numbers, you subtract their real parts and imaginary parts separately. For example, (3 + 2i) - (5 + 4i) = (3-5) + (2i-4i) = -2 - 2i.

5. How do you multiply and divide complex numbers?

To multiply complex numbers, you can use the FOIL method (First, Outer, Inner, Last). For example, (3 + 2i)(5 + 4i) = (3*5) + (3*4i) + (2i*5) + (2i*4i) = 15 + 12i + 10i + 8i^2 = 15 + 22i - 8 = 7 + 22i. To divide complex numbers, you can use the conjugate of the denominator to rationalize the fraction. For example, (3 + 2i) / (5 + 4i) = (3 + 2i)(5 - 4i) / (5 + 4i)(5 - 4i) = (15 - 12i + 10i - 8i^2) / (25 - 16i^2) = (23 - 2i) / 41.

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