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Suppose $x = 3 - 2i$ and $y = 4 + i$. Find both square roots of $y$. Then indicate which one is the principle square root.

Use the polar form of complex numbers to accomplish this task.

I'm not looking for an answer, as much as just some help as to how I would go about solving this problem. I really don't understand what I'm supposed to find.

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    $\begingroup$ What is the purpose of including $x$ here? Are you supposed to find the square roots of $x$ as well? $\endgroup$ – Arthur Dec 10 '14 at 16:50
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Write $y$ in polar form, and apply $\sqrt{re^{i\theta}}=\sqrt re^{i\theta/2}$ or $\sqrt{re^{i\theta}}=\sqrt re^{i(\pi+\theta/2)}$.

What has your teacher/book called "principle square root"?

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  • $\begingroup$ I believe he is asking to find the "east-most" root on the complex number plane. So the one in quadrant 1. $\endgroup$ – Jake Remmington Dec 10 '14 at 17:01
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To find square root of y you do not need x at all.

To find square root of a complex number take square root of its absolute value as modulus and only half of its argument. Second root is the conjugate, mirrored on real axis of Argand diagram.

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  • $\begingroup$ Can you show me what you are talking about here? I understand how to find the second root as the conjugate, but how do you find the first square root exactly? $\endgroup$ – Jake Remmington Dec 10 '14 at 17:08
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Suppose $z = \sqrt{y}$. We have $z^2 = y$, but we must take care!

For we can write $z$ in polar form as $z = r_z e^{i\theta_z}$. Now, when we compute $z^2$, we obtain:

$$z^2 = r_z^2 \left(e^{i\theta_z}\right)^2 = r_z^2 e^{2i\theta_z}.$$

Noting that complex numbers in polar form are only unique modulo $2\pi$, the presence of the $2$ in the exponent now reduces that uniqueness to modulo $\pi$. That means that two different values of $\theta_z$ can suffice!

First, we will compute $y$ in polar form:

$$r_y = \sqrt{4^2 + 1^2} = \sqrt{9} = 3.$$ $$\theta_y = \tan^{-1} \frac{1}{4} \approx 0.2449.$$

Next, we see that $$y = 3e^{0.2449i + 2i\pi k}, k \in \mathbb{Z}.$$

Setting this equal to $z^2$, we find

$$r_z = \sqrt{3}$$ $$\theta_z = \frac12\left(0.2449 + 2\pi k\right) = 0.1225 + \pi k.$$

Now, since any complex number is unique modulo $2\pi$, we see that

$$z = \sqrt{3}e^{0.1225i + i\pi k}$$

is the same number for all even values of $k$. It also takes a different value for odd values of $k$ (but the number is the same for all odd values of $k$).

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I implied that you use polar form. If you want to take square root of:

$ (a + i\, b) $ instead of $ (4 + i) $

$ ( a^2 + b^2)^ \frac12 \cdot $ Exp[$ i \cdot tan^{-1} (b/a)] $,

the answer is:

$ ( a^2 + b^2)^ \frac14 \cdot $ Exp[$ i \cdot tan^{-1} (b/a)/2] $

Now use Trig to get to a-b form.

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  • $\begingroup$ Be careful, as your values of $x,y$ are not what the OP used. $\endgroup$ – Emily Dec 10 '14 at 17:47

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