Simplify (-3i)(7i)(-1)

Complex number is the sum of a real number and an imaginary number. These are the numbers that can be written in the form of a+ib, where a and b both are real numbers. It is denoted by z.

 For example:

• 3+4i is a complex number, where 3 is a real number (Re) and 4i is an imaginary number (Im).
• 2+5i is a complex number where 2 is a real number (Re) and 5i is an imaginary number (im)

The combination of real number and imaginary number is called a Complex number

Example: √-3, √-7, √-11 are all imaginary numbers. here ‘i’ is an imaginary number called “iota”.

Rules of Imaginary Numbers

i = √-1

i² = -1

i³ = -i

i⁴ = 1

i⁴ⁿ = 1

i^4n-1 = -1

Simplify (-3i)(7i)(-1) 

Solution: 

Given: (-3i)(7i)(-1)

= -3i x 7i x (-1)

= -21i² x -1                            {i² = -1}

= -21 (-1) x -1 

= 21 x -1

= -21 + 0i

Similar Questions

Question 1: Express the problem in a+ib, (-5i)(1/8i)

Solution: 

Given: (-5i)(1/8i)

= -5/8 i²

= -5/8 (-1)                            {i² = -1}

=  5/8 + 0i

Question 2: Express the problem in a+ib, i⁹ + i¹⁹?

Solution:   

Given: i⁹ + i¹⁹

= (i ⁴ )²i + (i⁴ )⁴ i³

= (1)² i  + (1)⁴ i³                                {i⁴ = 1}

= i  + i³

= i + (-i)                   {i³ = -i}

= i – i 

= 0 + 0i

Question 3: Express in form of a+ib, {3(7+7i) + i(7+7i)}?

Solution: 

Given : {3(7+7i) + i(7+7i)}

= {21 +21i + 7i + 7i²}

= {21 + 28i + 7(-1)}

= {21 + 28i – 7}

= 14 + 28i

= 14 + 28i

Question 4: Express in form of a + bi, 1/(3-4i)?

Solution:  

Given : 1/(3-4i)

= 1/(3-4i)  x (3+4i)/(3+4i)

= (3+4i) / {(3)² – (4i)²}

= (3+4i) / {9 + 16}

= (3+4i) / 25

= 3/25 + (4/25)i

Question 5: Simplify (3 + 4i) / (3 + 2i)

Solution: 

Multiplying the numerator and denominator with the conjugate of denominators.

= ((3 + 4i) * (3 – 2i)) / ((3 + 2i) * (3 – 2i))

= (9 -6i +12i – 8i² ) / {9  -(2i)² }

= (17 + 6i) / (13)

= (17+ 6i) / 13

Question 6: Simplify (4 + 4i)*(3 – 4i)

Solution:   

Given : (4 + 4i)*(3 – 4i)

= (12+ 12i -16i – 16i² )

= 12 -4i +16

= 28 – 4i

Question 7: Simplify {(-3 – 5i) / (2 +2i)}?

Solution: 

Given {(-3 – 5i) / (2 +2i) }

conjugate of denominator 2+2i is 2-2i

Multiply the numerator and denominator with the conjugate

Therefore {(-3 – 5i) / (2 +2i) } x {(2-2i) / (2-2i) }

= {-6 -6i -10i +10i²  } /  { 2² – (2i)² }                  {difference of squares formula . i.e (a+b)(a-b) = a² – b² }

= {-6 -6i -10i + 10 (-1) } /  { 4 – 4(-1) }         { i² = -1 }

= {-6 -6i -10i -10 } / { 4 + 4 }

= (-16 – 16i ) / 8

= -16 /8  – 16i /8

= -2  -2i