# Class 11 RD Sharma Solutions – Chapter 13 Complex Numbers – Exercise 13.3

**Question 1. Find the square root of the following complex numbers.**

**(i) – 5 + 12i**

**(ii) -7 – 24i**

**(iii) 1 – i**

**(iv) – 8 – 6i**

**(v) 8 – 15i**

**(vi) **

**(vii) **

**(viii) 4i**

**(ix) -i**

**Solution:**

If b > 0,

If b < 0,

**(i) – 5 + 12i**

Given:

– 5 + 12i

We know, Z = a + ib

So,

Here, b > 0

Let us simplify now,

∴ Square root of (– 5 + 12i) is ±[2 + 3i]

**(ii) -7 – 24i**

Given:

-7 – 24i

We know, Z = -7 – 24i

So,

Here, b < 0

Let us simplify now,

∴ Square root of (-7 – 24i) is ± [3 – 4i]

**(iii) 1 – i**

Given:

1 – i

We know, Z = (1 – i)

So,

Here, b < 0

Let us simplify now,

∴ Square root of (1 – i) is ±

**(iv) -8 -6i**

Given:

-8 -6i

We know, Z = -8 -6i

So, = -8 -6i

Here, b < 0

Let us simplify now,

∴ Square root of (-8 -6i) is ± [1 – 3i]

**(v) 8 – 15i**

Given:

8 – 15i

We know, Z = 8 – 15i

So, = 8 – 15i

Here, b < 0

Let us simplify now,

∴ Square root of (8 – 15i) is ±

**(vi) **

Given:

We know, Z =

So,

= -11 – 60i

Here, b < 0

Let us simplify now,

∴ Square root of () is ± (5 – 6i)

**(vii)**

Given:

We know, Z =

So,

Here, b > 0

Let us simplify now,

∴ Square root of is ±

**(viii) 4i**

Given:

4i

We know, Z = 4i

So, = 4i

Here, b > 0

Let us simplify now,

∴ Square root of 4i is ±

**(ix) –i**

Given:

-i

We know, Z = -i

So, = -i

Here, b < 0

Let us simplify now,

∴ Square root of –i is ±

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