# Find the square root of (-16)

• Last Updated : 04 Apr, 2022

Complex numbers are terms that can be shown as the sum of real and imaginary numbers. 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. Here the value ‘a’ is called the real part which is denoted by Re(z), and ‘b’ is called the imaginary part Im(z) in form of a complex number.  It is also called an imaginary number. In complex number form a + bi ‘i’ is an imaginary number called “iota”. The value of i is (√-1) or we can write as i2 = -1. For example,

3 + 16i is a complex number, where 9 is a real number (Re) and 16i is an imaginary number (Im).15 + 20i is a complex number where 10 is a real number (Re) and  20i is an imaginary number (Im)

### Find the square root of (-16)

Solution:

As we have discussed above about the complex number

Now as per the question to find the square root of (-16).

= √-16

= √16(-1)

= (√16)(√-1)              {i = √-1}

= 4i

### Similar Questions

Question 1: Find the value of square root of {-25}?

Solution:

Given: √-25

= √-25

= √25(-1)

= (√25)(√-1)                   {i = √-1}

= 5i

Question 2: Find the value of square root of {-289}?

Solution:

Given : √-289

= √-289

= √289(-1)

= (√289)(√-1)                  {i = √-1}

= 17i

Question 3: Find the square of (9i)?

Solution:

After squaring an imaginary number, it gives a result in negative …

(9i)2 = 9i × 9i

= 81i2

= 81(-1)

= -100

Question 4: Simplify {(-3 – 5i) / (3 +2i)}?

Solution:

Given {(-3 – 5i) / (3 +2i)}

Conjugate of denominator 3+2i is 3-2i

Multiply with the conjugate of denominator

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

= {(-3-5i)(3-2i)} / {(3+2i)(3-2i)}

= {-9 +6i -15i +10i2 } / {32 – (2i)2}                  {difference of squares formula . i.e (a+b)(a-b) = a2 – b2 }

= {-9 + 6i -15i + 10 (-1)} / {9 – 4(-1)}         { i2 = -1 }

= {-9 + 6i -15i -10} / {9 + 4}

= (-19 – 9i) / 13

= -19 /13  – 9i /13

= -19/13 – 9/13 i

Question 5: Simplify the given expression 7/10i

Solution:

Given: 7/10i

Standard form of numerator, 7 = 7 +0i

Standard form of denominator, 10i = 0 +10i

Conjugate of denominator, 0 + 10i = 0 – 10i

Multiply the numerator and denominator with the conjugate,

Therefore, {(7 + 0i) / (0 + 10i)} × {(0 – 10i)/(0 – 10i)}

= {7(0 – 10i)} / {0 – (10i)2}

= {0 – 70i} / {0 – (100(-1))}

= {-70i} / 100

= 0 – 70/100i

= -7/10i

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