What is (-2)(2+3i)/(1+i)+(2-i)^2 ?

The solution to (-2)(2+3i)/(1+i)+(2-i)^2 is -2-5i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify (-2)(2+3i)/(1+i)+(2-i)^2

simplify

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

- 2 - 5 i

Solution steps

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

Apply rule:

(- a) = - a

(- 2) = - 2

= - 2 \cdot \frac{2 + 3 i}{1 + i} + (2 - i)^{2}

- 2 \cdot \frac{2 + 3 i}{1 + i} = - \frac{2 ( 2 + 3 i )}{1 + i}

(2 - i)^{2} = - 4 i + 3

= - \frac{2 ( 2 + 3 i )}{1 + i} - 4 i + 3

Simplify

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

= - 5 - i - 4 i + 3

Group like terms

= - i - 4 i - 5 + 3

Add similar elements:

- i - 4 i = - 5 i

= - 5 i - 5 + 3

Add the numbers:

- 5 + 3 = - 2

= - 5 i - 2

Rewrite in standard complex form:

- 2 - 5 i

= - 2 - 5 i

2 x + 1 < 1

3 x^{2} + x - 14 = 0

2 x^{2} - x = - 1

4 a^{2} - 9 x^{4}

793

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