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How can I find the square root of $z = −2i$ ? Am I going in the right direction by doing $-2i=a+bi$ ?

Riccardo Sven Risuleo
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Lisa Levi
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    Welcome to Mathematics Stack Exchange. You might do $-2i=(a+bi)^2$ or try polar coordinates. Note: $-2i$ has two square roots – J. W. Tanner Jun 19 '19 at 04:33
  • Polar co-ordinates are a useful tool for complex numbers and arise because of the identity $re^{i\theta}=r\cos \theta + ir\sin \theta$. You get two square roots because $1=1^2=(-1)^2$. Likewise there are three cube roots of $1$ etc - call one of these $\omega$ with $\omega \neq 1$ then if $a^3=z$ we also have $(\omega a)^3=(\omega^2 a)^3=z$ which identifies the other cube roots of $z$. – Mark Bennet Jun 19 '19 at 05:37
  • Well if $-2i = a + bi$ then $a=0$ and $b =-2$ and that just leaves you where you start. What you can do say the square root is $a+bi$ so $(a+bi)^2 = -2i$ so $(a+bi)^2 = a^2 +2abi + (bi)^2 = a^2 +2abi - b^2=(a^2 - b^2) + 2abi = -2i$. So $a^2-b^2=0$ and $2ab = -2$ and solving for $a$ and $b$. – fleablood Jun 19 '19 at 05:52
  • Possible duplicate of How do I get the square root of a complex number? – Hans Lundmark Jun 19 '19 at 06:33

3 Answers3

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$z = -2i = 2 e^{i3\pi/2}$. So, $\sqrt{z} = \pm \sqrt{2} e^{i3\pi/4} = \pm (1-i)$.

Vaneet
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My improved answer:

Let $\alpha=a+bi,$ with $a,b\in \Bbb R,$ and assume $-2i=\alpha^2=(a+bi)^2=\color{red}{a^2-b^2}+\color{green}{2ab}i.$

Therefore $\color{red}{a^2-b^2}=0$ and $\color{green}{ab}=-1.$

This means $\color{red}{a=\pm b}$. Because $ab=-1$, therefore, it must be that $a=-b$ and $\color{green}{a^2=1}.$

(It could not be that $a=b$ and $a^2=-1$, since $a\in\Bbb R$.)

Therefore $a=\pm1$ and $b=-a$; i.e., $\alpha=1-i$ or $-1+i$.

J. W. Tanner
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  • Let $\alpha=a+bi$ with $a,b\in \Bbb R$ and assume $-2i=\alpha^2=(a+bi)^2. $ Then $-4=(a+bi)^4=(a^4-6a^2b^2+b^4)+\color{blue}{(4a^3b-4ab^3)}i.$ This implies $\color{blue}{4ab(a+b)(a-b)}=0,$ so $a=0$ or $b=0$ or $a=\pm b.$ Now if $a=0$ then $\alpha=bi$ and $-2i=\alpha^2=-b^2$, a contradiction. Likewise if $b=0$ then $\alpha=a$ and $-2i=\alpha^2=a^2$, a contradiction. And if $a=b$ then $\alpha=a+ai=a(1+i)$ and $-2i=\alpha^2=a^2(1+i)^2=a^2(2i),$ a contradiction. So we're left with $a=-b,$ so $\alpha=a-ai=a(1-i)$ and $-2i=\alpha^2=a^2(1-i)^2=a^2(-2i),$ so $a^2=1$ so $a=\pm1$ and $\alpha=\pm(1-i)$ – J. W. Tanner Jun 19 '19 at 05:34
  • That comment (above) was my previous answer, but then I realized there was no need to calculate $(a+bi)^4$ – J. W. Tanner Jun 19 '19 at 05:35
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If you don't know polar coordinates:

We want to solve $z^2 = -2i = 0 + (-2)i$.

If $z = a+bi$ then $z^2 = (a+bi)^2$

$a^2 + 2abi + b^2i^2 =$

$a^2 + 2abi + b^2(-1) =$

$a^2 + 2abi - b^2=$

$(a^2 - b^2) + 2abi$.

And we know $z^2 = 0 + (-2i)$ so $(a^2 - b^2) + 2abi = 0 + (-2)i$.

So $a^2 - b^2 = 0$ and $2ab = -2$.

Solving for $a,b$ we have

$a^2 = b^2$ so $|a| = |b|$ and so $a = \pm b$. But $2ab = -2 < 0$ so $a$ and $b$ are opposite signs. So $a = -b$.

$2ab = -2$ so $ab = -1$. As $a=-b$ we have $(-b)b = -b^2 = -1$ so $b^2 =1$ so $b = \pm 1 $.

And $a = -b = \mp 1$.

So there are two square roots: $z = a + bi = -1 + i$ or $z = a+bi = 1 -i$.

.....

In general to solve $z^2 = u + v i$ we can solve by letting $z =a + bi$ and so $z^2 = (a^2 -b^2) + 2abi = u + vi$. SO we can solve

$a^2 - b^2 = u; 2ab = v$.

If $v \ne 0$ then $a \ne 0$ and $b = \frac v{2a}$ so

$a^2 - \frac {v^2}{4a^2} = u$

$4a^4 - v^2 = 4ua^2$ so

$4a^4 - 4ua^2 -v^2 = 0$ so

$a^2 = \frac {4u \pm \sqrt {16u^2 + 16v^2}}{8} = \frac {u \pm\sqrt{u^2 +v^2}}2$

But $a^2 \ge 0$ so $a^2 = \frac {u +\sqrt{u^2 +v^2}}2$

And $a = \pm \sqrt{\frac {u +\sqrt{u^2 +v^2}}2}$ and $b= \frac {v}{2a}$.

Best to learn polar coordinates. It' much easier.

fleablood
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