What is ((2+3i)(3-2i)-(2-3i)^2)/(17(1-i^{13))} ?

The solution to ((2+3i)(3-2i)-(2-3i)^2)/(17(1-i^{13))} is i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify ((2+3i)(3-2i)-(2-3i)^2)/(17(1-i^{13))}

simplify

\frac{( 2 + 3 i ) ( 3 - 2 i ) - ( 2 - 3 i ) ^{2}}{17 ( 1 - i ^{13} )}

i

Solution steps

\frac{( 2 + 3 i ) ( 3 - 2 i ) - ( 2 - 3 i ) ^{2}}{17 ( 1 - i ^{13} )}

Simplify

(2 + 3 i) (3 - 2 i) - (2 - 3 i)^{2} : 17 i + 17

= \frac{17 i + 17}{17 ( 1 - i ^{13} )}

Simplify

17 (1 - i^{13}) : 17 - 17 i

= \frac{17 i + 17}{17 - 17 i}

Factor

17 i + 17 : 17 (i + 1)

= \frac{17 ( i + 1 )}{17 - 17 i}

Factor

17 - 17 i : 17 (1 - i)

= \frac{17 ( i + 1 )}{17 ( 1 - i )}

Cancel the common factor:

17

= \frac{i + 1}{1 - i}

Simplify

\frac{i + 1}{1 - i} : i

= i

5 a^{2} - 45 b^{2}

∣ x ∣ = - 17

- 8 x - 19 < 21

\frac{x}{4} - \frac{x}{5} = \frac{3}{4}

9 f^{2} - 1

What is ((2+3i)(3-2i)-(2-3i)^2)/(17(1-i^{13))} ?
The solution to ((2+3i)(3-2i)-(2-3i)^2)/(17(1-i^{13))} is i