równanie zespolone

[lubierachowac]

\(\displaystyle{ \left| z\right|+z=8 +4i}\) (drugie z sprzężone)

[aalmond]

\(\displaystyle{ \sqrt{a^2 + b^2} + a - b \mbox{i} = 8 + 4 \mbox{i} \\
- b = 4}\)
itd.

[stuart clark]

Oryginał:    

\(\displaystyle{ \mid z \mid +z = 8+4i}\)

\(\displaystyle{ \mid z \mid -8 = -z+4i}\)

So here \(\displaystyle{ \text{L.H.S} =}\) Real number

then \(\displaystyle{ z=x+4i\;, x\in\mathbb{R}}\)

So \(\displaystyle{ \sqrt{x^2+16} = -x+8}\)

\(\displaystyle{ x^2+16 = (8-x)^2 = 64+x^2-16x}\)

\(\displaystyle{ 16x = 48\Leftrightarrow x=3}\)

So \(\displaystyle{ z=3+4i}\)

P.S = moderator sorry for not writting in polish

can you write into polish

\(\displaystyle{ \mid z \mid +z = 8+4i}\)

\(\displaystyle{ \mid z \mid -8 = -z+4i}\)

Lewa strona jest liczbą rzeczywistą, a więc \(\displaystyle{ z=x+4i\;, x\in\mathbb{R}}\)

czyli \(\displaystyle{ \sqrt{x^2+16} = -x+8}\)

\(\displaystyle{ x^2+16 = (8-x)^2 = 64+x^2-16x}\)

\(\displaystyle{ 16x = 48\Leftrightarrow x=3}\)

Zatem \(\displaystyle{ z=3+4i}\)