How do I find the solution to the equation $z^2=-81i$?

Question

This question is from the Powers of Complex Numbers, Precalculus section of KhanAcademy

Find the solution to the following equation whose argument is between $90°$ and $180°$

$$z^2=-81i$$

What I understand thus far:

I am going to set $r$ and $\theta$ to be the modulus and argument of $z$, respectively.

Therefore, $z^{ 2 }=r^{ 2 }[cos(2\cdot \theta )+isin(2\cdot \theta )]$

Now, I can understand how the modulus is $81$, but I do not understand how it was determined that the argument is $270°$ plus any multiple of $360°$. I am quite confused at this point and a hint in the right direction would be the best thing to help me figure out the solution to this problem and ones like it that I will encounter in the future.

There is also an algebraic way to find the square roots of a complex number, see this: http://math.stackexchange.com/questions/411174/square-roots-of-complex-numbers/411195#411195 And for your case, this one may also be useful: http://math.stackexchange.com/questions/3315/what-is-sqrti — Jean-Claude Arbaut, Mar 28 '16 at 10:04

Jan Eerland · Accepted Answer · 2016-03-28T10:07:42.833

2

$$z^2=-81i\Longleftrightarrow$$ $$z^2=|-81i|e^{\arg(-81i)i}\Longleftrightarrow$$ $$z^2=81e^{-\frac{\pi i}{2}}\Longleftrightarrow$$ $$z=\left(81e^{\left(2\pi k-\frac{\pi}{2}\right)i}\right)^{\frac{1}{2}}\Longleftrightarrow$$ $$z=9e^{\frac{1}{2}\left(2\pi k-\frac{\pi}{2}\right)i}$$

With $k\in\mathbb{Z}$ and $k:0-1$

So, the solutions are:

$$z_0=9e^{\frac{1}{2}\left(2\pi\cdot0-\frac{\pi}{2}\right)i}=9e^{-\frac{\pi i}{4}}$$ $$z_0=9e^{\frac{1}{2}\left(2\pi\cdot1-\frac{\pi}{2}\right)i}=9e^{\frac{3\pi i}{4}}$$

Now, notice that:

$$\color{red}{\frac{\pi}{2}<\arg\left[9e^{\frac{3\pi i}{4}}\right]<\pi\Longleftrightarrow\frac{\pi}{2}<\frac{3\pi}{4}<\pi}$$

So, your right answer is:

$$9e^{\frac{3\pi i}{4}}=9\cos\left(\frac{3\pi}{4}\right)+9\sin\left(\frac{3\pi}{4}\right)i=-\frac{9\sqrt{2}}{2}+\frac{9\sqrt{2}}{2}i$$

edited Mar 28 '16 at 10:07

answered Mar 28 '16 at 09:48

Jan Eerland

28,671

1

It seems that they will only accept the rectangular form like this $9cos(135°)+9sin(135°)\cdot i$. I am trying to figure out how they came up with $135°$. It does not seem very intuitive to me at all. – Cherry_Developer Mar 28 '16 at 09:55
@Cherry_Developer Notice that you can turn around in the complex plane and come were you started! – Jan Eerland Mar 28 '16 at 09:57
The hints to the problem on Khan Academy say: The argument of $-81i$ can be $270°$ plus any multiple of $360°$. This seems very arbitrary. I do not understand where they got this from. $270$ seems to have come out of thin air. – Cherry_Developer Mar 28 '16 at 10:03
No, it isn't because $\arg(-81i)=-\frac{\pi}{2}$ and $-\frac{\pi}{2}+2\pi=\frac{3\pi}{2}$ and that's equal to $270$ degrees! – Jan Eerland Mar 28 '16 at 10:05
Wait, is this because on an Argand diagram $-81i = 0 + -81i$, so that is just the diagram at $-90$ degrees? – Cherry_Developer Mar 28 '16 at 10:10
Hehe yes that's right! – Jan Eerland Mar 28 '16 at 10:11
Thank you so much!!! I felt a light bulb just go off in my head thanks to you! – Cherry_Developer Mar 28 '16 at 10:12
@Cherry_Developer You're welcome! I'm glad that I could help – Jan Eerland Mar 28 '16 at 10:13
@Jan Eerland \ Actually, $\mp\frac{9\sqrt{2}}2\pm\frac{9\sqrt{2}}2i\quad$. No? – Senex Ægypti Parvi Mar 28 '16 at 10:52
@SenexÆgyptiParvi Look at the question! – Jan Eerland Mar 28 '16 at 10:57
@Jan Eerland \ You are absolutely correct. Please disregard my comment or if you think it best, perhaps I ought to delete. It's your choice. – Senex Ægypti Parvi Mar 28 '16 at 11:02

score 0 · Answer 2 · answered Mar 28 '16 at 09:48

I think you have understood why the modulus is $81$.

Now, since $r^2(cos(2θ)+isin(2θ)) = -81i$ clearly, $[cos(2θ)]+[sin(2θ)]i = -i = (0) + (-1)i$.

Since the real and imaginary parts of the complex number are independent, you can equate them to get the answer.

So, you need to solve the equations $$cos(2\theta) = 0,\,\ sin(2\theta)=-1$$

score 0 · Answer 3 · answered Mar 28 '16 at 10:24

Write $-81i$ in trigonometric form: $$ -81i=81\cdot(-i)=81\left(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}\right) $$ so by De Moivre its square roots are $$ 9\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right) $$ and $$ 9\left(\cos\left(\frac{3\pi}{4}+\pi\right)+ i\sin\left(\frac{3\pi}{4}+\pi\right)\right) = 9\left(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4}\right) $$

Since $\pi/2<3\pi/4<\pi$, the first one is what you're looking for. Thus the answer is $$ 9\left(-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right) $$

If you use degrees, $\frac{3\pi}{2}=270^\circ$ and $\frac{3\pi}{4}=135^\circ$.

In general, if $z=r(\cos\alpha+i\sin\alpha)$, the $n$-th roots of $z$ are $$ \sqrt[n]{r}\left( \cos\left(\frac{\alpha}{n}+\frac{2k\pi}{n}\right) +i\sin\left(\frac{\alpha}{n}+\frac{2k\pi}{n}\right) \right), \qquad k=0,1,\dots,n-1 $$

score 0 · Answer 4 · answered Mar 28 '16 at 23:09

I would highly recommend using De Moivre's formula and graphing.

Start by getting some graphing paper and label your $x$-axis as real values from $-100$ to $100$ and label your $y$-axis as imaginary values from $-100i$ to $100i$. (go by $10$'s for the intervals.)

Place the complex number you are trying to square root on the graph.

Measure the angle it makes with the positive $x$-axis, the axis-line that points right. The angle measure for any real positive values is $0\deg$, naturally, and you can go from there? (Note: we measure our angles counterclockwise)

Now, all you have to do to square root this number is simply to 'half' the angle and square root the magnitude.

$$\text{magnitude}=|a+bi|=\sqrt{a^2+b^2}$$

In your case, $|-81i|=81$, so the square root of that is $9$.

Since we can see the angle measure is $270\deg$, half of that is $135\deg$. To calculate the exact value of that, the trigonometric equation is DeMoivre's Formula:

$$(a+bi)^n=|a+bi|^n\cdot(\cos(n\theta)+i\sin(n\theta))$$

For our case, $\sqrt{-81i}=(-81i)^{0.5}=81^{0.5}\cdot(\cos(0.5\cdot270)+i\sin(0.5\cdot270))=9(\cos(135)+i\sin(135))=9(-\frac{\sqrt2}{2}+i\frac{\sqrt2}2)$

Lastly, note that $270\deg=270+360\deg$ because the extra $360$ is just another spin around the circle, which ends up back at $270\deg$. However, when you half this 'new' angle, you get a different result, which notably explains why we sometimes have $x^2=a\implies x=\pm\sqrt a$ instead of $x=\sqrt a$.

If you want more insight on how DeMoivre's formula comes about, I recommend graphing $(1+i)^n$ for $n=0,1,2,3,\dots$ and measuring the angle measure and magnitude. (You can calculate $(1+i)^3=(1+i)(1+i)(1+i)$, for example. Just foiling)

How do I find the solution to the equation $z^2=-81i$?

4 Answers4