How can I solve for the roots of (z+i)^2=3-4i in my complex number homework?

In summary, the problem discussed is finding the roots of the equation (z+i)^2=3-4i, where z=3-4i and z^2=3-4i. The first part involves expanding z^2 to get the desired result, while the second part requires replacing z with 2-i and solving for the roots, which are z1=(2-i) and z2=-(2-i). There was also a mistake in the statement, as z should have been 2-i instead of 3-4i. Using x as the unknown variable may also lead to complex numbers.
  • #1
HMPARTICLE
95
0
1. The problem.

Given that z= 3-4i
Show that z^2 = 3-4i

Hence or otherwise find the roots of the equation (z+i)^2=3-4i

2. My attempt.

The first part of the problem is strait forward z^2= (2-i)(2-i) then expand to get the desired result.
Now the second part

(z+i)^2=3-4i. Becomes

z^2+ 2zi+i^2 = 3-4i

From here on I replace z with 2-i and get nowhere!
 
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  • #2
Given that z= 3-4i
leads to z² = -7-24i
There must be a mistake in your statement.

Besides that, the equation z²=3-4i has two solutions:

z1=(2-i) and z2=-(2-i)

Solving (z+i)²=3-4i is then straightforward.
 
  • #3
Yes there was a mistake! It should have read... Given that z = 2-i
 
  • #4
So you have (2 - i)2 = 3 - 4i and you are looking for roots for (x + i)2 = 3 - 4i (I've changed the unknown to x to avoid confusion).
 
  • #5
Yes! The variable z does confuse things a bit.
 
  • #6
HMPARTICLE said:
Yes! The variable z does confuse things a bit.

Don't forget, if you use x this way it is possibly a complex number !
 

FAQ: How can I solve for the roots of (z+i)^2=3-4i in my complex number homework?

1. What are complex numbers?

Complex numbers are numbers that have both a real part and an imaginary part. They are written in the form a + bi, where a and b are real numbers and i is the imaginary unit (√-1).

2. How do you add and subtract complex numbers?

To add or subtract complex numbers, you simply add or subtract the real parts and the imaginary parts separately. For example, (3 + 2i) + (1 + 4i) = (3 + 1) + (2i + 4i) = 4 + 6i. Similarly, (3 + 2i) - (1 + 4i) = (3 - 1) + (2i - 4i) = 2 - 2i.

3. How do you multiply complex numbers?

To multiply complex numbers, you can use the FOIL method (First, Outer, Inner, Last). For example, (3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8i² = (3 - 8) + (12 + 2)i = -5 + 14i.

4. How do you divide complex numbers?

To divide complex numbers, you need to multiply both the numerator and denominator by the complex conjugate of the denominator. The complex conjugate of a + bi is a - bi. For example, (3 + 2i) / (1 + 4i) = (3 + 2i)(1 - 4i) / (1 + 4i)(1 - 4i) = (3 - 12i + 2i - 8i²) / (1 - 16i²) = -5 - 10i / 17 = (-5/17) - (10/17)i.

5. Why are complex numbers important?

Complex numbers are important in mathematics and science because they can be used to solve equations that have no real solutions. They are also used in many applications, such as in electrical engineering, quantum mechanics, and signal processing. Additionally, complex numbers have many interesting and useful properties that make them a valuable tool in problem-solving.

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