Question

Simplify the radical and use the imaginary unit i.

Answer 1

Answer:

√

−

49

=

7

i

Explanation:

A square root of a number  

n

is a number  

x

such that  

x

2

=

n

Note that if  

x

is a Real number then  

x

2

≥

0

.

So any square root of  

−

49

is not a Real number.

In order to be able to take square roots of negative numbers, we need Complex numbers.

That's where the mysterious number  

i

comes into play. This is called the imaginary unit and has the property:

i

2

=

−

1

So  

i

is a square root of  

−

1

. Note that  

−

i

is also a square root of  

−

1

, since:

(

−

i

)

2

=

(

−

1

⋅

i

)

2

=

(

−

1

)

2

⋅

i

2

=

1

⋅

(

−

1

)

=

−

1

Then we find:

(

7

i

)

2

=

7

2

⋅

i

2

=

49

⋅

(

−

1

)

=

−

49

So  

7

i

is a square root of  

−

49

. Note that  

−

7

i

is also a square root of  

−

49

.

What do we mean by the square root of  

−

49

For positive values of  

n

, the square root is usually taken to mean the principal square root  

√

n

, which is the positive one.

For negative values of  

n

, the square roots are both multiples of  

i

, so neither positive nor negative, but we can define:

√

n

=

i

√

−

n

With this definition, the principal square root of  

−

49

is:

√

−

49

=

i

√

49

=

7

i

Footnote

The question remains: Where does  

i

come from?

It is possible to define Complex numbers formally, as pairs of Real numbers with rules for arithmetic like this:

(

a

,

b

)

+

(

c

,

d

)

=

(

a

+

c

,

b

+

d

)

(

a

,

b

)

⋅

(

c

,

d

)

=

(

a

c

−

b

d

,

a

b

+

c

d

)

These rules for addition and multiplication work as expected with commutativity, distributivity, etc.

Then Real numbers are just Complex numbers of the form  

(

a

,

0

)

and we find:

(

0

,

1

)

⋅

(

0

,

1

)

=

(

−

1

,

0

)

That is  

(

0

,

1

)

is a square root of  

(

−

1

,

0

)

Then we can define  

i

=

(

0

,

1

)

and:

(

a

,

b

)

=

a

(

1

,

0

)

+

b

(

0

,

1

)

=

a

+

b

i