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In an ODE class, one assignment question says find the “rectangular” expression z = a + bi (with a and b real) and the “polar” expression |z|, Arg(z) where z is "a square root of i with negative imaginary part".

The answer is given as $(-1 - i) / \sqrt{2}$ for rectangular, and r=1, $\theta = 5\pi / 4$ for polar. I am aware of Euler's formula, the powers of complex numbers, etc. This answer however did not make sense to me.

Any ideas?

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2  
$\mathrm{i}=\mathrm{e}^{\mathrm{i}\pi/2}$. –  Did Jul 12 '11 at 10:58

3 Answers 3

up vote 2 down vote accepted

The two square roots of a complex $w$ with module $\left\vert w\right\vert $ and $\theta =\arg (w)$, i.e $w=\left\vert w\right\vert e^{i\theta }$ are $$ \sqrt{w}=\sqrt{\left\vert w\right\vert }e^{i\frac{\theta +2k\pi }{2}},k\in \{0,1\}.$$

For $w=i$, we have $\left\vert w\right\vert =\left\vert i\right\vert =1$, $% \theta =\arg (i)=\frac{\pi }{2}$. Thus $$\sqrt{e^{i\pi /2}}=e^{i\frac{\pi /2+2k\pi }{2}},k\in \{0,1\}.$$ One of the roots is

$$z_{1}=e^{i\frac{\pi /2}{2}}=e^{i\pi /4}=\cos \left( \frac{\pi }{4}\right) +i\sin \left( \frac{\pi }{4}\right) =\frac{1}{2}\sqrt{2}+\frac{1}{2}i\sqrt{2} $$ and the other $$z_{2}=e^{i\frac{\pi /2+2\pi }{2}}=e^{i5\pi /4}=\cos \left( \frac{5}{4% }\pi \right) +i\sin \left( \frac{5}{4}\pi \right) =-\frac{1}{2}\sqrt{2}-% \frac{1}{2}i\sqrt{2}.$$

Only $z_{2}$ has negative imaginary part. Observing that $\frac{1}{2}\sqrt{2}% =\frac{1}{\sqrt{2}}$, we get $$z_{2}=\frac{1}{\sqrt{2}}\left( -1-i\right) .$$

Added 2: $w=i,z_1,z_2$

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Added: The $n^{th}$ roots of the complex $w=\left\vert w\right\vert e^{i\theta }$ are

$$\sqrt[n]{w}=\sqrt[n]{\left\vert w\right\vert }e^{i\frac{\theta +2k\pi }{n}},k\in \left\{ 0,1,2,\ldots ,n-1\right\} ,$$

because

$$\begin{eqnarray*} \left( \sqrt[n]{w}\right) ^{n} &=&\left( \sqrt[n]{\left\vert w\right\vert }% e^{i\frac{\theta +2k\pi }{n}}\right) ^{n}=\left( \sqrt[n]{\left\vert w\right\vert }\right) ^{n}\left( e^{i\frac{\theta +2k\pi }{n}}\right) ^{n} \\ &=&\left\vert w\right\vert e^{i\left( \theta +2k\pi \right) }=\left\vert w\right\vert e^{i\theta }e^{i2k\pi }=\left\vert w\right\vert e^{i\theta }\cdot 1 \\ &=&\left\vert w\right\vert e^{i\theta }=w. \end{eqnarray*}$$

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It is clearest to use polar coordinates first. If $z = re^{i\theta}$ (polar form of a complex number), then the $n^{th}$ power of $z$ is simply $z^n = r^ne^{in\theta}$. Now, to solve $z^2 = i$: Since $i = e^{i\pi/2}$, we are looking for $z = re^{i\theta}$ such that $r^2 = 1$ and $2\theta = \pi/2$. Well if we take $r = 1$, then $\theta = \pi/4$ OR $\theta = 5\pi/4$ (since $5\pi/2$ is the same angle measure as $\pi/2$). A little thought shows there are no other distinct complex numbers that work, so $z = e^{i\pi/4}$ and $z = e^{i5\pi/4}$ are the square roots of $i$.

Then use Euler's formula to express these in rectangular, and you find that $e^{i5\pi/4} = (-1 - i)/\sqrt{2}$, having negative imaginary part.

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Another way is to say $(a+bi)^2=i$ and separate into real and imaginary parts. Then $a^2-b^2=0,\ 2ab=1,\ a^2-\frac{1}{4a^2}=0,\ 4a^4=1,\ a=\pm\frac{1}{\sqrt{2}}, b=\pm\frac{1}{\sqrt{2}}$. Then trying the four possibilities, we find the two square roots of $i$ are $\frac{\pm (1+ i)}{\sqrt{2}}$ and you can take the minus sign.

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+1 for this nice way. –  Américo Tavares Jul 13 '11 at 11:07

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