Question

solve the system of equation
Re(z^2)=0,|z|=2

Answer 1

\(a=a+bi\)
\[Re(z^2)=0\iff a^2-b^2=0\iff a=\pm b\]

Answer 2

oops i meant
\[z=a+bi\]

Answer 3

then
\[z^2=a^2-b^2+2abi\] making
\[Re(z^2)=a^2-b^2\]

Answer 4

if \(a^2-b^2=0\) the \(a=b\) or \(a=-b\) so the number lives either along the line \(a=b\) or along the line \(a=-b\)

Answer 5

if \(|z|=2\) it also lives on the circle of radius \(2\)

Answer 6

is it clear what those four numbers are?

Answer 7

what r they?

Answer 8

what would they be if it was the unit circle rather than the circle of radius 2?

Answer 9

take those four answers and double them

Answer 10

hint, if it was the unit circle, since one angle is clearly \(\frac{\pi}{4}\) the one in quadrant 1 would be \[\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\]

Answer 11

I contribute to the satellite73's stuff, that is \(\pm a=\pm b\), so that you have 4 possibilities:
a =b
a=-b
-a = b
-a=-b
if |z|=2, it means |z| = \(\sqrt{a^2+b^2}\) =2
all above possibility are the same when we compute a^2 and b^2, so that just take a =b as sample
a=b--> a^2+b^2 = 2a^2 --> |z| = \(\sqrt{2a^2}=2\) if and only if \(a^2 =2 --> a=\pm\sqrt{2}\)
If \(a=\sqrt{2}\) and a = b, then z= a +bi = \(\sqrt{2}+\sqrt{2}i= \sqrt{2}(1+i)\)\)
If \(a=\sqrt{2}\) and a =- b, then z= a +bi = \(-\sqrt{2}+\sqrt{2}i= \sqrt{2}(-1+i)\)\)
do the same with the 2 possibilities left ( I mean -a=b and -a=-b), you will get the required answer