Find the square root of complex number i - Maths - Complex Numbers and Quadratic Equations

Question

Find the square root of complex number i

Pronay · Accepted Answer

1)

Let a +ib be the square root of i then

$\sqrt{i} = a + i b$

Squaring both sides

$i = {(a + i b)}^{2} i = a^{2} + i^{2} b^{2} + 2 i a b i = a^{2} - b^{2} + 2 i a b$

Comparing the coefficients of i from both sides of equation we get

$2 a b = 1 ...... (1)$

Comparing the coefficients of constant from both sides of equation we get

$a^{2} - b^{2} = 0 ...... (2)$

Substituting the value of b from equation (1) to (2)

$a^{2} = {(\frac{1}{2 a})}^{2} 4 a^{4} = 1 \Rightarrow a^{2} = \pm \frac{1}{2}$

Since a and b are real , therefore we reject $a^{2} = - \frac{1}{2}$

Hence

$a^{2} = \frac{1}{2}$

Therefore

$a^{2} = \frac{1}{2} \Rightarrow a = \pm \frac{1}{\sqrt{2}}$

and equation (1) implies

$b = \frac{1}{2} a = \frac{1}{2} \times \pm \frac{1}{\sqrt{2}} \Rightarrow b = \pm \frac{1}{2 \sqrt{2}}$

Therefore roots of i are

$\frac{1}{\sqrt{2}} + i \frac{1}{2 \sqrt{2}} and - \frac{1}{\sqrt{2}} - i \frac{1}{2 \sqrt{2}}$

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