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Question Number 107690 by bemath last updated on 12/Aug/20

        “BeMath“  Let the complex number z satisfies the  equation 3(z−1)= i(z+1)   (1) find z in the form a+bi where a,b ∈R   (2) find the value of ∣z∣ and ∣z−z^∗ ∣

$$\:\:\:\:\:\:\:\:``\mathcal{B}{e}\mathcal{M}{ath}`` \\ $$$${Let}\:{the}\:{complex}\:{number}\:{z}\:{satisfies}\:{the} \\ $$$${equation}\:\mathrm{3}\left({z}−\mathrm{1}\right)=\:{i}\left({z}+\mathrm{1}\right)\: \\ $$$$\left(\mathrm{1}\right)\:{find}\:{z}\:{in}\:{the}\:{form}\:{a}+{bi}\:{where}\:{a},{b}\:\in\mathbb{R}\: \\ $$$$\left(\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\mid{z}\mid\:{and}\:\mid{z}−{z}^{\ast} \mid\: \\ $$$$ \\ $$

Answered by mr W last updated on 12/Aug/20

(1)  z=a+bi  3(a+bi−1)=i(a+bi+1)  (3a−3)+3bi=−b+(a+1)i  3a−3=−b ⇒3a+b=3  3b=a+1 ⇒a−3b=−1  9a+3b=9  ⇒a=(4/5)  ⇒b=(3/5)

$$\left(\mathrm{1}\right) \\ $$$${z}={a}+{bi} \\ $$$$\mathrm{3}\left({a}+{bi}−\mathrm{1}\right)={i}\left({a}+{bi}+\mathrm{1}\right) \\ $$$$\left(\mathrm{3}{a}−\mathrm{3}\right)+\mathrm{3}{bi}=−{b}+\left({a}+\mathrm{1}\right){i} \\ $$$$\mathrm{3}{a}−\mathrm{3}=−{b}\:\Rightarrow\mathrm{3}{a}+{b}=\mathrm{3} \\ $$$$\mathrm{3}{b}={a}+\mathrm{1}\:\Rightarrow{a}−\mathrm{3}{b}=−\mathrm{1} \\ $$$$\mathrm{9}{a}+\mathrm{3}{b}=\mathrm{9} \\ $$$$\Rightarrow{a}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\Rightarrow{b}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$

Commented by bemath last updated on 12/Aug/20

∣z∣=(√(((16)/(25))+(9/(25)))) = 1 . it correct?

$$\mid{z}\mid=\sqrt{\frac{\mathrm{16}}{\mathrm{25}}+\frac{\mathrm{9}}{\mathrm{25}}}\:=\:\mathrm{1}\:.\:{it}\:{correct}?\: \\ $$

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