What is ((2-i)^2-i^{15})/(3-i) ?

The solution to ((2-i)^2-i^{15})/(3-i) is 6/5-3/5 i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify ((2-i)^2-i^{15})/(3-i)

simplify

\frac{( 2 - i ) ^{2} - i ^{15}}{3 - i}

\frac{6}{5} - \frac{3}{5} i

Solution steps

\frac{( 2 - i ) ^{2} - i ^{15}}{3 - i}

Multiply by the conjugate

\frac{3 + i}{3 + i}

= \frac{( ( 2 - i ) ^{2} - i ^{15} ) ( 3 + i )}{( 3 - i ) ( 3 + i )}

Simplify

\frac{( ( 2 - i ) ^{2} - i ^{15} ) ( 3 + i )}{( 3 - i ) ( 3 + i )} : \frac{- 6 i + 12}{10}

= \frac{- 6 i + 12}{10}

Rewrite in standard complex form:

\frac{6}{5} - \frac{3}{5} i

= \frac{6}{5} - \frac{3}{5} i

(y^{2})^{5}

(- 2)^{2} \cdot (- 2)^{3} \cdot (- 2)^{4}

2 (x^{2} - 1) + 3 x (x - 4)

\frac{1}{cos ( θ ) sin ( θ )}

\frac{1}{50}

What is ((2-i)^2-i^{15})/(3-i) ?
The solution to ((2-i)^2-i^{15})/(3-i) is 6/5-3/5 i