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,

0]\\ =\pm\left[\left(\frac{-5+\sqrt{25+144}}{2}\right)^{\frac{1}{2}}+i\left(\frac{5+\sqrt{25+144}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{-5+\sqrt{169}}{2}\right)^{\frac{1}{2}}+i\left(\frac{5+\sqrt{169}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{-5+13}{2}\right)^{\frac{1}{2}}+i\left(\frac{5+13}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{8}{2}\right)^{\frac{1}{2}}+i\left(\frac{18}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[4^{\frac{1}{2}}+i9^{\frac{1}{2}}\right]\\ =\pm[2+3i]" title="Rendered by QuickLaTeX.com" height="399" width="863" style="vertical-align: -7px;">

∴ 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,

0]\\ =\pm\left[\left(\frac{1+\sqrt{1+48}}{2}\right)^{\frac{1}{2}}+i\left(\frac{-1+\sqrt{1+48}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{1+\sqrt{49}}{2}\right)^{\frac{1}{2}}+i\left(\frac{-1+\sqrt{49}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{1+7}{2}\right)^{\frac{1}{2}}+i\left(\frac{-1+7}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{8}{2}\right)^{\frac{1}{2}}+i\left(\frac{6}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[4^{\frac{1}{2}}+i3^{\frac{1}{2}}\right]\\ =\pm[2+\sqrt3i]" title="Rendered by QuickLaTeX.com" height="432" width="864" style="vertical-align: -7px;">

∴ Square root of  is ± 

(viii) 4i

Given:

4i

We know, Z = 4i

So,  = 4i

Here, b > 0

Let us simplify now,

0]\\ =\pm\left[\left(\frac{0+\sqrt{0+16}}{2}\right)^{\frac{1}{2}}+i\left(\frac{0+\sqrt{0+16}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{0+\sqrt{16}}{2}\right)^{\frac{1}{2}}+i\left(\frac{0+\sqrt{16}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{0+4}{2}\right)^{\frac{1}{2}}+i\left(\frac{0+4}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{4}{2}\right)^{\frac{1}{2}}+i\left(\frac{4}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[2^{\frac{1}{2}}+i2^{\frac{1}{2}}\right]\\ =\pm[\sqrt2+\sqrt2i]\\ =\pm\sqrt2(1+i)" title="Rendered by QuickLaTeX.com" height="436" width="735" style="vertical-align: -7px;">

∴ 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 ±