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Exercise 2.9 · Q6

Q.If zz is a non zero complex number, such that 2iz2=z‾2iz^2=\overline z then ∣z∣|z| is

(1) 12\dfrac12
(2) 11
(3) 22
(4) 33
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Taking moduli converts the equation into a real equation purely in ∣z∣|z|, which is quicker than solving for x,yx,y separately and then computing x2+y2\sqrt{x^2+y^2}.

Step 1. Take the modulus of both sides of 2iz2=z‾2iz^2=\overline z.

∣2iz2∣=∣z‾∣.|2iz^2|=|\overline z|.

Step 2. Simplify each side using ∣ab∣=∣a∣∣b∣|ab|=|a||b|, ∣zn∣=∣z∣n|z^n|=|z|^n, and ∣z‾∣=∣z∣|\overline z|=|z|.

∣2∣∣i∣∣z∣2=∣z∣  ⟹  2⋅1⋅∣z∣2=∣z∣  ⟹  2∣z∣2=∣z∣.|2||i||z|^2=|z| \;\Longrightarrow\; 2\cdot1\cdot|z|^2=|z| \;\Longrightarrow\; 2|z|^2=|z|.

Step 3. Since z≠0z\ne0, we have ∣z∣≠0|z|\ne0; divide both sides by ∣z∣|z|.

2∣z∣=1  ⟹  ∣z∣=12.2|z|=1 \;\Longrightarrow\; |z|=\frac12. …

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