Let z and w be complex numbers such that |2z - w| = 25, |z + 2w| = 5, and |z + w| = 2. Find |z|.

I need a little help with absolute values on complex numbers.

You asked for it:

2Z-W=(2Zx-Wx)i+(2Zy-Wy)j
|2Z-W|^2=4Zx^2-4ZxWx+Wx^2+4Zy-4ZyWy+Wy^2
=4|Z|^2+|W|^2 -4(ZxWx+ZyWy)= 625
and
|Z+2W|^2 =|Z|^2+4|W|^2 +4(ZxWx+ZyWy)=25
so
625+25 =5[ |Z|^2+|W|2| ]
or
130 = |Z|^2 + |W|^2
now that last one
|Z+W|^2= Zx^2+2ZxWx+Wx^2+Zy^2+2ZyWy+Wy^2] = 4
or
|Z|^2+|W|^2 +2(ZxWx+ZyWy)=4
or
ZxWx+ZyWy=-63
go back up
625 = 4|Z|^2+|W|^2+4*63
25 = |Z|^2+4|W|^2 -4*63
------------------------
or
2500 = 16|Z|^2 + 4 |W|^2 +16(63)
25 = |Z|^2 + 4 |W|^2 -4(63)
------------------------------
2475 = 15 |Z|^2 + 20(63)
all yours :)

at least one typo

2Z-W=(2Zx-Wx)i+(2Zy-Wy)j
|2Z-W|^2=4Zx^2-4ZxWx+Wx^2+4Zy-4ZyWy+Wy^2
=4|Z|^2+|W|^2 -4(ZxWx+ZyWy)= 625
and
|Z+2W|^2 =|Z|^2+4|W|^2 +4(ZxWx+ZyWy)=25
so
625+25 =5[ |Z|^2+|W|^2| ]
or
130 = |Z|^2 + |W|^2
now that last one
|Z+W|^2= Zx^2+2ZxWx+Wx^2+Zy^2+2ZyWy+Wy^2] = 4
or
|Z|^2+|W|^2 +2(ZxWx+ZyWy)=4
or
ZxWx+ZyWy=-63
go back up
625 = 4|Z|^2+|W|^2+4*63
25 = |Z|^2+4|W|^2 -4*63
------------------------
or
2500 = 16|Z|^2 + 4 |W|^2 +16(63)
25 = |Z|^2 + 4 |W|^2 -4(63)
------------------------------
2475 = 15 |Z|^2 + 20(63)

Questions that may come in exams about complex numbers

mathematics Foundation

To solve this problem, we need to use the properties of absolute values and complex numbers. The absolute value of a complex number z, denoted as |z|, represents the distance between the origin (0, 0) and the point representing the complex number z in the complex plane.

Let's start by using the information given to find the value of |z|. We know that |z + w| = 2. This means that the distance between z and -w (the additive inverse of w) is 2.

To simplify the problem, we can consider z + w as a new complex number u = z + w. Therefore, |u| = 2. Now we need to find the value of |z|.

We can use the properties of absolute values to simplify the expression |2z - w| = 25. We can rewrite this as |2(z + w) - 3w| = 25.

If we let v = 2u - 3w, the equation becomes |v| = 25. Rearranging the equation, we have |2u - 3w| = 25, which means the distance between 2u and 3w is 25.

We also have |u + w| = 5. Using the same logic as before, we let x = u + w, so |x| = 5. This gives us the distance between u and -w is 5.

Now we have three equations:

|u| = 2
|v| = 25
|x| = 5

Let's solve these equations one by one.