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Question: If \(z_{1} \neq z_{2}\) and\(|z_{1}| = |z_{2}|\) (where\(z_{1}\), then \(z_{2}\) is....

If z1≠z2z_{1} \neq z_{2} and∣z1∣=∣z2∣|z_{1}| = |z_{2}| (wherez1z_{1}, then z2z_{2} is.

A

(z1+z2)(z1−z2)\frac{(z_{1} + z_{2})}{(z_{1} - z_{2})}

B

zz

C

z4+z+2=0z^{4} + z + 2 = 0

D

∣z∣<1|z| < 1

Answer

(z1+z2)(z1−z2)\frac{(z_{1} + z_{2})}{(z_{1} - z_{2})}

Explanation

Solution

∵\because

arg(z)=θ=tan⁡−1yxarg(z) = \theta = \tan^{- 1}\frac{y}{x}

arg(z‾)=θ=tan⁡−1(−yx)arg(\overline{z}) = \theta = \tan^{- 1}\left( \frac{- y}{x} \right)

arg(z)≠a⥂rg(z‾)arg(z) \neq a ⥂ rg(\overline{z})

∴ ∣z∣=4|z| = 4.