Mathematics, 02.09.2020 05:01 id0617045

Evaluate the expression and write your answer in the form a + bi. 1) (2-6i)+(4+2i)2) (6+5i)(9-2i)3) 2/(3-9i)4) (3 − 5i)(7 − 2i)

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Answer from: Alex9089435028

Simplified form of the given expressions will be,

1). 6 - 4i

2). 64 + 33i

3). $\frac{1}{15}+\frac{1}{5}i$

4). $\frac{31}{53}-\frac{31}{53}i$

1). In a complex number (a + bi),

a = Real part of the complex number

bi = Imaginary part

Expression given in the question → (2 - 6i) + (4 + 2i)

Rule to solve the given expression,

"Add real part and imaginary part of two complex numbers separately"

(2 - 6i) + (4 + 2i) = (2 + 4) + (-6i + 2i)

= 6 + (-4i)

= 6 - 4i

Therefore, simplified form of the given expression will be (6 - 4i).

2). Given expression → (6 + 5i)(9 - 2i)

Rule to solve the expression → i² = -1

(6 + 5i)(9 - 2i) = 6(9 - 2i) + 5i(9 - 2i)

= 54 - 12i + 45i - 10i²

= 54 - 12i + 45i + 10 [Since, i² = -1]

= (54 + 10) + (45i - 12i)

= 64 + 33i

Therefore, simplified form of the given expression will be (64 + 33i).

3). Given expression → $\frac{2}{3-9i}$

Multiply numerator and denominator with the conjugate of (3 - 9i) to convert the expression into (a + bi).

$\frac{2}{3-9i}= \frac{2(3+9i)}{(3-9i)(3+9i)}$

$=\frac{2(3+9i)}{3^2-(9i)^2}$

$=\frac{6+18i}{9-81(-1)}$

$=\frac{6+18i}{9+81}$

$=\frac{6+18i}{90}$

$=\frac{6(1+3i)}{90}$

$=\frac{1}{15}+\frac{1}{5}i$

Therefore, simplified form of the given expression will be $\frac{1}{15}+\frac{1}{5}i$

4). Given expression → $\frac{(3-5i)}{(7-2i)}$

Multiply numerator and denominator with the conjugate of the denominator.

$\frac{(3-5i)}{(7-2i)}=\frac{(3-5i)(7+2i)}{(7-2i)(7+2i)}$

$=\frac{3(7+2i)-5i(7+2i)}{7^2-(2i)^2}$

$=\frac{21+4i-35i-10i^2}{49-4(-1)}$

$=\frac{21-31i+10}{53}$

$=\frac{31-31i}{53}$

$=\frac{31}{53}-\frac{31}{53}i$

Therefore, simplified form of the given expression will be $\frac{31}{53}-\frac{31}{53}i$ .

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Answer from: tomtom6870

Step-by-step explanation:

Given the following complex numbers, we are to expressed them in the form of a+bi where a is the real part and b is the imaginary part of the complex number.

1) (2-6i)+(4+2i)

open the parenthesis

= 2-6i+4+2i

collect like terms

= 2+4-6i+2i

= 6-4i

2) (6+5i)(9-2i)

= 6(9)-6(2i)+9(5i)-5i(2i)

= 54-12i+45i-10i²

= 54+33i-10i²

In complex number i² = -1

= 54+33i-10(-1)

= 54+33i+10

= 54+10+33i

= 64+33i

3) For the complex number 2/(3-9i), we will rationalize by multiplying by the conjugate of the denominator i.e 3+9i

= 2/3-9i*3+9i/3+9i

=2(3+9i)/(3-9i)(3+9i)

= 6+18i/9-27i+27i-81i²

= 6+18i/9-81(-1)

= 6+18i/9+81

= 6+18i/90

= 6/90 + 18i/90

= 1/15+1/5 i

4) For (3 − 5i)(7 − 2i)

open the parenthesis

= 3(7)-3(2i)-7(5i)-5i(-2i)

= 21-6i-35i+10i²

= 21-6i-35i+10(-1)

= 21-41i-10

= 11-41i

Answer from: Quest

The answer is a hope i

Answer from: Quest

can i see the following ?

step-by-step explanation:

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