Solve for b if $1 - \frac{b}{4 + i \sqrt{3}} + \frac{b}{4 - i \sqrt{3}} = 0$ $1-b/(4+isqrt3)+b/(4-isqrt3)=0$ ?

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Answer 1 · 2017-12-16T01:24:35

$b = \frac{19 i \sqrt{3}}{6}$ $b=(19isqrt3)/6$

$1 - \frac{b}{4 + i \sqrt{3}} + \frac{b}{4 - i \sqrt{3}} = 0$ $1-b/(4+isqrt3)+b/(4-isqrt3)=0$

$1 - \frac{b \cdot (4 - i \sqrt{3}) - b \cdot (4 + i \sqrt{3})}{16 - (- 3)} = 0$ $1-[b*(4-isqrt3)-b*(4+isqrt3)]/(16-(-3))=0$

$1 - \frac{- 2 b i \sqrt{3}}{19} = 0$ $1-(-2bisqrt3)/19=0$

$\frac{2 b i \sqrt{3}}{19} = - 1$ $(2bisqrt3)/19=-1$

$b = - \frac{19}{2 i \sqrt{3}}$ $b=-19/(2isqrt3)$

$b = \frac{- 19 i \sqrt{3}}{- 6}$ $b=(-19isqrt3)/(-6)$

$b = \frac{19 i \sqrt{3}}{6}$ $b=(19isqrt3)/6$

Solve for *b* if 1-b/(4+isqrt3)+b/(4-isqrt3)=01−b4+i√3+b4−i√3=01-b/(4+isqrt3)+b/(4-isqrt3)=0?