FP1 Complex Numbers

Is the question asking you to solve the simultaneous equations to find $z$ and $w \ ?$

Here's my working out simplified

z=-iw + 4 + 4i [1]

z^2 - w^2 = -2 + 16i [2]

Sub [1] into [2]:

( -iw + 4 + 4i )( -iw + 4 + 4i ) - w^2 = -2 + 16i

-2w^2 - 8iw + 8w + 32i = -2 + 16i

w^2 + 4wi - 4w = 1 - 8i



Thanks a lot for any help

Here's my working out simplified

z=-iw + 4 + 4i [1]

z^2 - w^2 = -2 + 16i [2]

Sub [1] into [2]:

( -iw + 4 + 4i )( -iw + 4 + 4i ) - w^2 = -2 + 16i

-2w^2 - 8iw + 8w + 32i = -2 + 16i

w^2 + 4wi - 4w = 1 - 8i



Thanks a lot for any help

Yeah, but even with that how would I go about finding the complex numbers w and z?
w^2 + 4iw - 4w = 1 + 8i

Equating real parts:
w^2 - 4w=1
w^2 - 4w - 1=0
w=2±√5
but this wouldn't be a complex number as there's no i there.

Also when equating the imaginary parts, I get 4w=8 so w=2. This isn't a complex number. Oh I'm so confused: (

Can you post the full question? I'll be able to give you a definitive answer if I see it.

Solve the pair of simultaneous equations
z^2 - w^2 = -2 + 16i
z + iw = 4 + 4i

If z+iw=4+4i
what is z-iw?
If you can write that down then you can get z^2+w^2 and hence solve with z^2-w^2=-2+16i to find z^2 and w^2.

z+iw=4+4i
z-iw=4-4i
Squaring both sides gives
z^2 - w^2 - 2zwi = -32i

You already know what z^2 - w^2 is - they tell you in the question

What do you get if you multiply a complex number by its conjugate?

If z+iw=4+4i
what is z-iw?
If you can write that down then you can get z^2+w^2 and hence solve with z^2-w^2=-2+16i to find z^2 and w^2.

Thanks for this reminder.



z+iw=4+4i
z-iw=4-4i

(z-iw)(z+iw)=z^2 + w^2

z^2 + w^2 = (4-4i)(4+4i)
z^2 + w^2 = 16 - 16i^2

z^2 + w^2 = 32 [1]
z^2 - w^2 = -2 + 16i [2] ----this is given in question

[1] - [2]:
2w^2 = 34 - 16i

w^2 = 17 - 8i

Sub w^2 into [2]

z^2 = 15 - 8i

So how does this help us get just z and w?

Thanks

find the squareroots

How do I do that?

this is what I tried
z^2 = 15 - 8i
Let z=a + bi
(a+bi)(a+bi)=15-8i
a^2 - b^2 + 2abi=15-8i
Real: a^2 - b^2 = 15 [1]
Imaginary: 2ab=-8, so a=-4/b [2]

Sub[2] into [1]:
(-4/b)^2 - b^2 = 15
b^4 -1=0
(b^2 +1)(b^2 -1)=0
Since b is an element of the real number set,
b=±1

So a=±4

So z=-4 + i or z=4- i

What did I do wrong here? I'd continue to do something similar to get w, but these complex numbers for z are wrong

The working you've done is correct so you must've made a mistake beforehand or the ans is wrong.

How do I do that?

this is what I tried
z^2 = 15 - 8i
Let z=a + bi
(a+bi)(a+bi)=15-8i
a^2 - b^2 + 2abi=15-8i
Real: a^2 - b^2 = 15 [1]
Imaginary: 2ab=-8, so a=-4/b [2]

Sub[2] into [1]:
(-4/b)^2 - b^2 = 15
b^4 -1=0
(b^2 +1)(b^2 -1)=0
Since b is an element of the real number set,
b=±1

So a=±4

So z=-4 + i or z=4- i

What did I do wrong here? I'd continue to do something similar to get w, but these complex numbers for z are wrong

(4-i)^2=15-8i so this answer is correct

But my textbook says that
z=2+3i, w=1-2i
z=2+I, w=3-2i

Is the book wrong?

But my textbook says that
z=2+3i, w=1-2i
z=2+I, w=3-2i

Is the book wrong?