What is 5-3*(z-z^{-1})/(2*i) ?

The solution to 5-3*(z-z^{-1})/(2*i) is 5+(3z^2-3)/(2z)i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify 5-3(z-z^{-1})/(2i)

simplify

5 - 3 \cdot \frac{z - z ^{- 1}}{2 \cdot i}

5 + \frac{3 z ^{2} - 3}{2 z} i

Solution steps

5 - 3 \cdot \frac{z - z ^{- 1}}{2 i}

Simplify

3 \cdot \frac{z - z ^{- 1}}{2 i} : - \frac{3 i ( z ^{2} - 1 )}{2 z}

= 5 - (- \frac{3 i ( z ^{2} - 1 )}{2 z})

Join

5 - (- \frac{3 i ( z ^{2} - 1 )}{2 z}) : \frac{10 z + 3 z ^{2} i - 3 i}{2 z}

= \frac{10 z + 3 z ^{2} i - 3 i}{2 z}

Rewrite

10 z + 3 z^{2} i - 3 i

in standard complex form:

10 z + (3 z^{2} - 3) i

= \frac{10 z + ( 3 z ^{2} - 3 ) i}{2 z}

Apply the fraction rule:

\frac{a + bi}{c} = \frac{a}{c} + \frac{b}{c} i

= \frac{10 z}{2 z} + \frac{3 z ^{2} - 3}{2 z} i

Cancel

\frac{10 z}{2 z} : 5

= 5 + \frac{3 z ^{2} - 3}{2 z} i

\frac{2}{1 + 2}

11 x + 11 = - 2 x - 8

b^{2} + 1 1^{2} = 2 1^{2}

- 10 + 3 x = 5

y, 3 x + 5 y = 50

What is 5-3*(z-z^{-1})/(2*i) ?
The solution to 5-3*(z-z^{-1})/(2*i) is 5+(3z^2-3)/(2z)i