Square Root of a Complex Number & Solving Complex Equations

Let z1 = x1 + iy1 be the given complex number and we have to obtain its square root.

Let x + iy = (x1 + iy1)½

Squaring ,

=> x2 – y2 + 2ixy = x1 + iy1

=> x1 = x2 – y2 and y1 = 2 xy

=> x2 – y12 /4x2 = x1

$\large x^2 = \frac{|z_1 | + x_1}{2} $

$\large y = \pm \frac{y_1}{|y_1|} = \sqrt{\frac{|z_1|-x_1}{2}}$

If y1 > 0   ; then ,

$\large x = \pm \sqrt{\frac{|z_1|+x_1}{2}} \; , \; y = \pm \sqrt{\frac{|z_1|-x_1}{2}} $

$\large \sqrt{x_1 + i y_1} = \pm (\sqrt{\frac{ |z_1| + Re(z_1) }{2}} + i \sqrt{\frac{ |z_1| – Re(z_1) }{2}})$

If y1 < 0 ; then ,

$\large x = \pm \sqrt{\frac{|z_1|+x_1}{2}} \; , \; y = \mp \sqrt{\frac{|z_1|-x_1}{2}} $

$\large \sqrt{x_1 + i y_1} = \pm (\sqrt{\frac{ |z_1| + Re(z_1) }{2}} – i \sqrt{\frac{ |z_1| – Re(z_1) }{2}}) $

Illustration : Find the square root of 8 – 15 i

Solution: Here y = -15 < 0

$\large \sqrt{8-15 i} = \pm (\sqrt{\frac{|z| + Re(z) }{2}} – i \sqrt{\frac{ |z| – Re(z) }{2}})$

$\large \pm \frac{1}{\sqrt{2}} (5-3 i)$

An Alternate Method to find the square root : 

(i) If the imaginary part is not even then multiply and divide the given complex number by 2.

e.g.  z = 8 –15 i , here imaginary part is not even so write

z = (16 – 30 i) and Let a + i b =16– 30 i.

(ii) Now divide the numerical value of imaginary part of a + ib by 2 and let quotient be P and find all possible two factors of the number P thus obtained and take that pair in which difference of squares of the numbers is equal to the real part of a + ib.

e.g. here numerical value of Im(16 – 30 i) is 30 .

Now 30 = 2 × 15. All possible way to express 15 as a product of two are 1 × 15, 3 × 5, etc.
here 52 – 32 = 16 = Re (16 – 30i) so we will take 5, 3.

(iii) Take i with the smaller or the greater factor according as the real part of a + ib is positive or negative and if real part is zero then take equal factors of P and associate i with any one of them.

e.g. Re(16 – 30 i) > 0, we will take i with 3.

Now complete the square and write down the square root of z. e.g.

$\large z = \frac{1}{2}[16-30 i]$

$\large = \frac{1}{2}[5^2 + (3 i)^2 – 2 \times 5 \times 3 i]$

$\large = \frac{1}{2}[5-3 i]^2$

$\large \sqrt{z} = \pm \frac{1}{\sqrt{2}} (5-3 i)$

Solving Complex Equations :

Simple equations in z may be solved by putting z = a + ib in the equation and equating the real part on the L.H.S with the real part on the R.H.S and the imaginary part on the L.H.S with the imaginary part on the R.H.S.

Illustration : Solve for z i.e., find all complex numbers z which satisfy |z|2 – 2iz + 2c(1+ i) =0 where c is real.

Solution: Put z = a + ib then a2 + b2 – 2ai + 2b + 2c + 2ci = 0

⇒ (a2 + b2 + 2b + 2c) + (2c – 2a) i = 0

⇒  a2 + b2 + 2b + 2c = 0 and 2c – 2a = 0

⇒ a = c

Now b2 + 2b + (c2 + 2c) = 0

⇒ $\large b = -1 \pm \sqrt{1-2c-c^2} $

Since b is real , 1- 2c – c2 ≥ 0

⇒ c ∈ [ -1 – √2 , -1 + √2 ]

⇒ $\large z = c + i (-1 \pm \sqrt{1-2c-c^2}) $

Solution of the given equation does not exits for

c ∉ [ -1 – √2 , -1 + √2 ]

Conjugate of a Complex Number:

The conjugate of the complex number z = a + ib is defined as
a – ib and is denoted by $\bar{z}$  .

In other words z is the mirror image of z in the real axis.

If z = a + ib, $\large  z + \bar{z} = 2 a (Real)$

$\large  z – \bar{z} = 2 i b (Imaginary)$

and , $\large  z . \bar{z} = ( a + ib)(a – ib) = a^2 + b^2 (Real) $

$\large = |z|^2 = | \bar{z}|^2  $

Also , $\large Re(z) = \frac{z + \bar{z}}{2} $

$\large Im(z) = \frac{z – \bar{z}}{2 i} $

Exercise: 

(i) Find the real θ such that $\large \frac{3 + 2 i sin\theta }{1-2 i sin\theta} $ is

(a) Real (b) Purely Imaginary

(ii) Simplify $\large \frac{20}{\sqrt{3} – \sqrt{-2}} + \frac{30}{3 \sqrt{-2} – 2\sqrt{3}} – \frac{14}{2 \sqrt{3} – \sqrt{-2}}$

(iii) Find the square root of :

(a) -8 -6 i

(b) $\large \frac{x^2}{y^2} + \frac{y^2}{x^2} – \frac{1}{i}(\frac{x}{y} – \frac{y}{x} ) – \frac{9}{4} $

Also Read :

Basic Concepts , Modulus and Argument of a Complex Number
Geometrical meaning of addition , subtraction , multiplication & division
Properties of Conjugate , Modulus & Argument
De Moivre’s Theorem & Applications of De Moivre’s Theorem
Concept of Rotation in Complex Number
Condition for common root(s)

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