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Question: For any two complex numbers \(z.\bar{z} = |\bar{z}|^{2}\)we have \(\overline{z_{1} + z_{2}} = \overl...

For any two complex numbers z.zˉ=∣zˉ∣2z.\bar{z} = |\bar{z}|^{2}we have z1+z2‾=z1‾+z2‾\overline{z_{1} + z_{2}} = \overline{z_{1}} + \overline{z_{2}} argz=argzˉargz = arg\bar{z} then.

A

∣z∣=4|z| = 4

B

a⥂rgz=5π6,a ⥂ rgz = \frac{5\pi}{6},

C

23−2i2\sqrt{3} - 2i

D

23+2i2\sqrt{3} + 2i

Answer

∣z∣=4|z| = 4

Explanation

Solution

We have 4i4=i\frac{4i}{4} = i

⇒ (z)=π/2[∵tan⁡θ=b/a](z) = \pi/2\lbrack\because\tan\theta = b/a\rbrack

Where z=−21+3iz = \frac{- 2}{1 + \sqrt{3}i}

⇒ −21+3i×1−3i1−3i\frac{- 2}{1 + \sqrt{3}i} \times \frac{1 - \sqrt{3}i}{1 - \sqrt{3}i}

⇒ =−2+23i1+3= \frac{- 2 + 2\sqrt{3}i}{1 + 3}

Note : Also ⇒z=−12+32i\Rightarrow z = \frac{- 1}{2} + \frac{\sqrt{3}}{2}i

⇒ ⇒arg(z)=tan⁡−1(−3/21/2)=2π3\Rightarrow arg(z) = \tan^{- 1}\left( - \frac{\sqrt{3}/2}{1/2} \right) = \frac{2\pi}{3} is purely imaginary.