
How do you multiply $ (i)(2i)( - 7i) $ ?
Answer
531.6k+ views
Hint: The complex number consists of the real part and an imaginary part and is denoted by “Z”. It can be expressed as $ z = a + ib $ where “a” is the real part and “b” is the imaginary part. Here we will take the given expression and multiply the real terms with real and imaginary “I” with “I”
Complete step-by-step answer:
Take the given expression: $ (i)(2i)( - 7i) $
Now, multiply constant terms with constants and variables with variable by using power and exponent property –
$ = - 14{i^3} $
Now, by using the properties of imaginary “I”
$ {i^3} = - i, $ since $ {i^2} = - 1 $
$ = - 14( - i) $
Product of two negative term gives positive term
$ = 14i $
This is the required solution.
So, the correct answer is “ $ = 14i $ ”.
Note: Know the concepts of the complex numbers. It is the combination of real numbers and the imaginary numbers and since imaginary numbers are very difficult to understand and therefore, they are complex numbers. To find the modulus of the complex number be good in squares and square-root. Also, be good in multiples and simplifications of the equation. Remembering the square of the negative terms also gives the positive values.
Complete step-by-step answer:
Take the given expression: $ (i)(2i)( - 7i) $
Now, multiply constant terms with constants and variables with variable by using power and exponent property –
$ = - 14{i^3} $
Now, by using the properties of imaginary “I”
$ {i^3} = - i, $ since $ {i^2} = - 1 $
$ = - 14( - i) $
Product of two negative term gives positive term
$ = 14i $
This is the required solution.
So, the correct answer is “ $ = 14i $ ”.
Note: Know the concepts of the complex numbers. It is the combination of real numbers and the imaginary numbers and since imaginary numbers are very difficult to understand and therefore, they are complex numbers. To find the modulus of the complex number be good in squares and square-root. Also, be good in multiples and simplifications of the equation. Remembering the square of the negative terms also gives the positive values.
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