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# 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 ± 