How do you find the square root of an imaginary number of the form a+bi?

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How do you find the square root of an imaginary number of the form a+bi?

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Answer 1 · 2015-12-24T13:19:25

Alternatively, solve without using trigonometry to find that the square roots of $a + b i$ $a+bi$ are:

$\pm ⎛ ⎝ ⎛ ⎝ \sqrt{\frac{\sqrt{a^{2} + b^{2}} + a}{2}} ⎞ ⎠ + ⎛ ⎝ \frac{b}{| b |} \sqrt{\frac{\sqrt{a^{2} + b^{2}} - a}{2}} ⎞ ⎠ i ⎞ ⎠$ $+-((sqrt((sqrt(a^2+b^2)+a)/2)) + (b/abs(b) sqrt((sqrt(a^2+b^2)-a)/2))i)$

Explanation:

Suppose $a + b i = {(c + d i)}^{2}$ $a+bi = (c+di)^2$

How do we solve for $c$ $c$ and $d$ $d$ ?

${(c + d i)}^{2} = c^{2} + 2 c d i + d^{2} i^{2} = (c^{2} - d^{2}) + (2 c d) i$ $(c+di)^2 = c^2+2cdi + d^2i^2 = (c^2-d^2) + (2cd)i$

So we want to solve:

$c^{2} - d^{2} = a$ $c^2-d^2 = a$

$2 c d = b$ $2cd = b$

From the second of these, we find:

$d = \frac{b}{2 c}$ $d = b/(2c)$

So:

$d^{2} = \frac{b^{2}}{4 c^{2}}$ $d^2 = b^2/(4c^2)$

Substituting this in the first equation we get:

$c^{2} - \frac{b^{2}}{4 c^{2}} = a$ $c^2-b^2/(4c^2) = a$

Multiply through by $4 c^{2}$ $4c^2$ to get:

$4 {(c^{2})}^{2} - b^{2} = 4 a c^{2}$ $4(c^2)^2-b^2 = 4ac^2$

Subtract $4 a c^{2}$ $4ac^2$ from both sides to get:

$4 {(c^{2})}^{2} - 4 a (c^{2}) - b^{2} = 0$ $4(c^2)^2-4a(c^2)-b^2 = 0$

From the quadratic formula, we find:

$c^{2} = \frac{4 a \pm \sqrt{{(4 a)}^{2} + 16 b^{2}}}{8} = \frac{a \pm \sqrt{a^{2} + b^{2}}}{2}$ $c^2 = (4a+-sqrt((4a)^2+16b^2))/8 =(a+-sqrt(a^2+b^2))/2$

For $c$ $c$ to be Real valued we require $c^{2} \geq 0$ $c^2 >= 0$ , hence we need to choose the root with the $+$ $+$ sign...

$c^{2} = \frac{a + \sqrt{a^{2} + b^{2}}}{2}$ $c^2 = (a+sqrt(a^2+b^2))/2$

Hence:

$c = \pm \sqrt{\frac{\sqrt{a^{2} + b^{2}} + a}{2}}$ $c = +-sqrt((sqrt(a^2+b^2)+a)/2)$

Then:

$d = \pm \sqrt{c^{2} - a}$ $d = +-sqrt(c^2-a)$

$= \pm \sqrt{\frac{\sqrt{a^{2} + b^{2}} + a}{2} - a}$ $= +-sqrt((sqrt(a^2+b^2)+a)/2-a)$

$= \pm \sqrt{\frac{\sqrt{a^{2} + b^{2}} - a}{2}}$ $= +-sqrt((sqrt(a^2+b^2)-a)/2)$

The remaining question is: What signs do we need to choose?

Since $2 c d = b$ $2cd = b$ :

If $b > 0$ $b > 0$ then $c$ $c$ and $d$ $d$ must have the same signs.

If $b < 0$ $b < 0$ then $c$ $c$ and $d$ $d$ must have opposite signs.

If $b = 0$ $b=0$ then $d = 0$ $d=0$ , so we don't have to worry.

If $b \neq 0$ $b != 0$ then we can use $\frac{b}{| b |}$ $b/abs(b)$ as a multiplier to match the signs as we require to find that the square roots of $a + b i$ $a+bi$ are:

$\pm ⎛ ⎝ ⎛ ⎝ \sqrt{\frac{\sqrt{a^{2} + b^{2}} + a}{2}} ⎞ ⎠ + ⎛ ⎝ \frac{b}{| b |} \sqrt{\frac{\sqrt{a^{2} + b^{2}} - a}{2}} ⎞ ⎠ i ⎞ ⎠$ $+-((sqrt((sqrt(a^2+b^2)+a)/2)) + (b/abs(b) sqrt((sqrt(a^2+b^2)-a)/2))i)$

Footnote

The question asked what is the square root of $a + b i$ $a+bi$ .

For positive Real numbers $x$ $x$ , the principal square root of $x$ $x$ is the positive one, which is the one we mean when we write $\sqrt{x}$ $sqrt(x)$ . This is also the square root that people commonly mean when they say "the square root of $2$ $2$ " and suchlike, neglecting the fact that $2$ $2$ has two square roots.

For negative Real numbers $x$ $x$ , then by convention and definition, $\sqrt{x} = i \sqrt{- x}$ $sqrt(x) = i sqrt(-x)$ , That is, the principal square root is the one with a positive coefficient of $i$ $i$ .

It becomes more complicated when we deal with the square roots of Complex numbers in general.

Consider:

$\sqrt{- \frac{1}{2} - \frac{\sqrt{3}}{2} i} = \pm (\frac{1}{2} - \frac{\sqrt{3}}{2} i)$ $sqrt(-1/2-sqrt(3)/2i) = +-(1/2-sqrt(3)/2 i)$

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How do you find the square root of an imaginary number of the form a+bi?

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