Complex Number Help: Find x+yi for z1,z1z2,z1/z2

[Kris1]

z1 = −11+2 i and z2 = −1+13 i are given

I need to find the following in the form x + y i.

conjugate of z1 =

conjugate of z1z2=

z1/z2 =

a) how would i go about it
and
b) can someone provide the solutions to the questions if possible

 

[mathworker]

z1 = −11+2 i and z2 = −1+13 i are given

I need to find the following in the form x + y i.

/z1 =

/z1z2=

z1/z2 =

a) how would i go about it
and
b) can someone provide the solutions to the questions if possible

a) first of all you have to rationalize the function \(\displaystyle 1/(x+iy)\)...that means you you should remove the i term from \(\displaystyle x+iy\) in the denominator ... do you now how to do that?...multiply it with \(\displaystyle x-iy\) and see the result

 

[Prove It]

z1 = −11+2 i and z2 = −1+13 i are given

I need to find the following in the form x + y i.

/z1 =

/z1z2=

z1/z2 =

a) how would i go about it
and
b) can someone provide the solutions to the questions if possible

The first two "things" you need to do are unreadable, please fix them.

 

[Nono713]

I have got just got the solution to the third one. However i can't work out the solution to the first two conjugate equations. Can you provide the answer to the first one? or at least how to solve it

Assuming you mean $\frac{1}{z_1}$, multiply both sides of the fraction by the complex conjugate of the denominator $z_1$. This'll give you a real denominator and you can split the fraction into real and imaginary parts then.

Consider:

$$\frac{1}{x + iy} = \frac{1}{x + iy} \cdot \frac{x - iy}{x - iy} = \frac{x - iy}{(x + iy)(x - iy)} = \frac{x - iy}{x^2 + y^2} = \frac{x}{x^2 + y^2} - i \frac{y}{x^2 + y^2}$$

And $x^2 + y^2$ is real by definition. Does it make sense now?

 

[Prove It]

z1 = −11+2 i and z2 = −1+13 i are given

I need to find the following in the form x + y i.

conjugate of z1 =

conjugate of z1z2=

z1/z2 =

a) how would i go about it
and
b) can someone provide the solutions to the questions if possible

Now that you have made your questions readable, you should know that the conjugate of a complex number is defined as having the same real part and the negative imaginary part of the original complex number.

 

[Kris1]

So u are saying that for the first one my answer should be -11+(-2*i)? is that what i am understanding?

 

[Prove It]

So u are saying that for the first one my answer should be -11+(-2*i)? is that what i am understanding?

Yes :)

 

[mathworker]

Yes :)

i think answer is \(\displaystyle -11-2i\) divided by its modulus

 

[Prove It]

i think answer is \(\displaystyle -11-2i\) divided by its modulus

Well you think wrong. If \(\displaystyle \displaystyle z = x + i\,y \) then \(\displaystyle \bar{z} = x - i\,y \). You do not divide by its modulus.

You are instead thinking of finding a unit vector in the same direction as an original vector.

 

[mathworker]

Well you think wrong. If \(\displaystyle \displaystyle z = x + i\,y \) then \(\displaystyle \bar{z} = x - i\,y \). You do not divide by its modulus.

You are instead thinking of finding a unit vector in the same direction as an original vector.

sorry, i read the question as \(\displaystyle 1/z1\)

 

[Prove It]

sorry, i read the question as \(\displaystyle 1/z1\)

I hope I didn't offend you, it's not your fault that there were many interpretations of the original post :)

 

[TheBigBadBen]

z1 = −11+2 i and z2 = −1+13 i are given

I need to find the following in the form x + y i.

conjugate of z1 =

conjugate of z1z2=

z1/z2 =

a) how would i go about it
and
b) can someone provide the solutions to the questions if possible

I realize that with the other posts, this will be somewhat redundant, but I hope it will be useful nevertheless to have everything consolidated.

The Question:
Given the numbers $z_1 = -11 + 2i$ and $z_2 = -1 + 13i$, find:
i) the conjugate of $z_1$ (which I'll denote $\overline{z_1}$)
ii) $\overline{z_1 \, z_2}$
iii) $\overline{z_1/z_2}$

The Solution:

We begin by noting that for any complex number in the form $z=a+bi$, we may simply write $\overline{z}=a-bi$. Thus, the answer to i) is simply
$$\overline{z_1}=-11-2i$$
The second bit is a little trickier, because we have to multiply two complex numbers. There are different methods to use here, but I always prefer to "FOIL" it, that is, expand it as the usual product of binomials, remembering that $i^2=1$:
$$z_1\,z_2=(-11 + 2i)(-1 + 13i)=11-22i-143i+26i^2=11-26-22i-143i\\
=-15-165i$$
From there, it's easy to find the conjugate of the result:
$$\overline{z_1\,z_2}=-15+165i$$
See what I did there?
Another way we could have done this is by finding the conjugate of each number, then multiplying. In other words:
$$\overline{z_1\,z_2}=(\overline{z_1})(\overline{z_2})$$
As you will find out for problems more intricate than this, this becomes important later.

Part 3 is tricker still, because now we're dividing fractions. I'll make that its own post

 

[TheBigBadBen]

Part 3:
We now need to find $z_1/z_2$. Because having a complex number in the denominator is tricky, we are going to rationalize it. That is, we write:
$$\frac{z_1}{z_2}
=\frac{z_1}{z_2}\cdot \frac{\overline{z_2}}{\overline{z_2}}
\\=\frac{z_1\,\overline{z_2}}{z_2\,\overline{z_2}}
\\=\frac{z_1\,\overline{z_2}}{\left| z_2 \right| ^2}$$
That is, in order to get rid of the complex number in the denominator, we multiply by the entire fraction by the conjugate of the denominator over itself. Since the product of any complex number with its conjugate yields the square of its modulus, we now have a real number in the denominator, making our lives much easier.

So, we have:
$$\frac{−11+2i }{−1+13i }
=\frac{−11+2i}{-1+13i}\cdot \frac{-1-13i}{-1-13i}
\\=\frac{(-11+2i)\,(-1-13i)}{(-1+13i)\,(-1-13i)}$$

From here, simply expand the product as we did before. You should end up with

$$=\frac{37+141i}{170}\\
=\frac{37}{170}+\frac{141}{170}i
$$

Now, we simply need to take the conjugate, so flip the second term to get
$$\overline{\left(\frac{z_1}{z_2}\right)}=
\frac{37}{170}-\frac{141}{170}i$$