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z-3-2i-1-i-z-z-




Question Number 11355 by uni last updated on 21/Mar/17
z+3−2i=(1+i)×z^−  ⇒∣z∣=?
$$\mathrm{z}+\mathrm{3}−\mathrm{2i}=\left(\mathrm{1}+\mathrm{i}\right)×\overset{−} {\mathrm{z}}\:\Rightarrow\mid\mathrm{z}\mid=? \\ $$
Answered by sou1618 last updated on 22/Mar/17
set a,b: R    z=a+bi ,z^� =a−bi    ∣z∣=(√(a^2 +b^2 ))    (a+bi)+3−2i=(1+i)(a−bi)  a+bi+3−2i=a+b+ai−bi  3−2i=b+(a−2b)i  so     { ((3=b)),((−2i=(a−2b)i)) :}⇒ { ((a=4)),((b=3)) :}  so    z=4+3i    ∣z∣=5
$${set}\:{a},{b}:\:\mathbb{R} \\ $$$$\:\:{z}={a}+{bi}\:,\bar {{z}}={a}−{bi} \\ $$$$\:\:\mid{z}\mid=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$ \\ $$$$\left({a}+{bi}\right)+\mathrm{3}−\mathrm{2}{i}=\left(\mathrm{1}+{i}\right)\left({a}−{bi}\right) \\ $$$${a}+{bi}+\mathrm{3}−\mathrm{2}{i}={a}+{b}+{ai}−{bi} \\ $$$$\mathrm{3}−\mathrm{2}{i}={b}+\left({a}−\mathrm{2}{b}\right){i} \\ $$$${so} \\ $$$$\:\:\begin{cases}{\mathrm{3}={b}}\\{−\mathrm{2}{i}=\left({a}−\mathrm{2}{b}\right){i}}\end{cases}\Rightarrow\begin{cases}{{a}=\mathrm{4}}\\{{b}=\mathrm{3}}\end{cases} \\ $$$${so} \\ $$$$\:\:{z}=\mathrm{4}+\mathrm{3}{i} \\ $$$$\:\:\mid{z}\mid=\mathrm{5} \\ $$

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