If z and w are complex numbers such that $|z+w|$ = $|z-w|$, prove that $\arg(z)-\arg(w)= \pm\pi/2$. Can this be solved algebraically or would a graphic interpretation be better. Both methods woul
. z bar in complex numbers
in your solution, you find, correctly, a and b such that 3+4i 1−i + 2−i 2+3i = a + bi. but you need to find the modulus and the argument of the number. That is, you need to find r > 0 and θ ∈ [0, 2π) such that. 3 + 4i 1 − i + 2 − i 2 + 3i = r(cos θ + i sin θ). There is a simple way to converting between the standard a + bi format
And I've turned z $\overline{w}$ into polar form which gave me 1cis $\frac{\pi}{3}$ so I know that that lies on unit circle too (obviously anyway cause z and w lie on it anyway) From what I know when multiplying complex numbers I can just add their $\theta$, so I need two that add to $\frac{\pi}{3}$ which is quite a few of numbers.
3. Find the number of complex numbers satisfying |z| = z + 1 + 2i | z | = z + 1 + 2 i . My method: I know |z| | z | is real. So, the imaginary part of the RHS should be equal to 0 0. So, z z should be of the form x − 2i x − 2 i . Using that I am getting an imaginary value for x itself!