check if i separated real and imaginary parts correctly

[sleek bluff]

Pls check and in img below

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[sleek bluff]

https://cdn.discordapp.com/attachments/692048202253205635/1129853718431539290/1689448501133.jpg

[long owl]

Well, 2 is obviously wrong. You can't cancel like that.

You need to multiply top and bottom by 2 - 3i.

[sleek bluff]

Quick question, how do I know when I need to multiply equation by 'bottom' ?

[long owl]

When you have a complex number in the denominator, usually.

[sleek bluff]

Thankyou

Are others correct?

[long owl]

Also, note that both real and imaginary part are real functions. Imaginary part doesn't contain i by itself.
So, z = Re(z) + i Im(z), not z = Re(z) + Im(z).
So, for example, in 1 we need to have Re(z^2) = x^2 - y^2, Im(z^2) = 2xy.

As for the others, let's see.

Actually, let me try myself, looks pretty fun.

[sleek bluff]

Gotcha , when writing imaginary parts as "imaginary part = 2xy", don't write i in answer, only put 'i' in equations

[long owl]

Yes.

[sleek bluff]

So in example 1, real part = x²-y² ?

[sleek bluff]

Another related question: does sign matter when multiplying equation by bottom? Why reverse sign?

[long owl]

z^2 = (x + iy)^2 = (x^2 - y^2) + 2ixy
Re(z^2) = x^2 - y^2
Im(z^2) = 2xy
2.
3i/(2 + 3i) = 3i(2 - 3i)/(4 + 9) = (9 + 6i)/13
Re(3i/(2 + 3i)) = 9/13
Im(3i/(2 + 3i)) = 6/13
3.
I'll denote the conjugate of z as z'.
iz + 2z' = i(x + iy) + 2x - 2iy = -y + ix + 2x - 2iy = (2x - y) + (x - 2y)i
Re(iz + z') = 2x - y
Im(iz + z') = x - 2y
4.
z^2 + z' = x^2 - y^2 + 2ixy + x - iy = (x^2 - y^2 + x) + (2xy - y)i
Re(z^2 + z') = x^2 - y^2 + x
Im(z^2 + z') = 2xy - y

[long owl]

Well, obviously. Only zz' will be a real number, so if you see a + bi, you multiply by a - bi.

[sleek bluff]

Ok, just multiply with switched signs, got it

[long owl]

Yeah.

[sleek bluff]

+close