z-z-1-3-i-find-

Question Number 207563 by hardmath last updated on 18/May/24

z + ∣ z ∣ = 1 + \sqrt{3} i

find : φ = ?

Answered by Frix last updated on 18/May/24

z + ∣ z ∣= 1 + \sqrt{3} i

∣ z ∣= 1 + \sqrt{3} i - z

∣ z ∣\in R \Rightarrow z = a + \sqrt{3} i

∣ a + \sqrt{3} i ∣= 1 - a

\sqrt{a^{2} + 3} = 1 - a \Rightarrow a = - 1

z = - 1 + \sqrt{3} i = {2 e}^{i \frac{2 π}{3}} \Rightarrow φ = \frac{2 π}{3}

Commented by hardmath last updated on 19/May/24

thank you professor

Answered by A5T last updated on 18/May/24

z = a + b i \Rightarrow∣ z ∣= \sqrt{a^{2} + b^{2}}

z + ∣ z ∣= (a + \sqrt{a^{2} + b^{2}}) + b i = 1 + i \sqrt{3} \Rightarrow b = \sqrt{3}

a + \sqrt{a^{2} + b^{2}} = 1 \Rightarrow a + \sqrt{a^{2} + 3} = 1 \Rightarrow a^{2} + 3 = 1 + a^{2} - 2 a

\Rightarrow a = - 1 \Rightarrow z = - 1 + i \sqrt{3}

\Rightarrow z = 2 e^{i \frac{2 π}{3}}

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