square root of a complex number [Solved!]

X

I cant get a formula for the square root of `a + bi` to work.

Relevant page

<a href="http://stanleyrabinowitz.com/bibliography/complexSquareRoot.pdf">http://stanleyrabinowitz.com/bibliography/complexSquareRoot.pdf</a>

What I've done so far

I start with  `sqrt(a+ b i),` translate into polar or exponential (`re^(i θ)`) coordinates and back again, and get 

`sqrt(a+b i) = sqrt(r) (cos [theta] + i  sin [theta]).`

where 

`theta = 0.5 arcsin (b/r)`

`r =sqrt( (a^2 +b^2)`

Then I test   this with `(2 + 3 i )^2 = -5 + 12 i.`

I get  

`theta = 0.5 arcsin(12/13) = 0.5880`

`sqrt(-5 + 12 i)` ` = sqrt(13)( cos[0.5880 ] + i sin [0.5880]).`

which turns out to be 

` sqrt( 13) ( (.83)+ i (.01) )` 

or about 

`3 + i 0.036` 

which is ridiculous. What is my error ?

Thanks, 

Jedothek

X

@Jedothek: I formatted your question so it was easier to read. I encourage you to make use of the "add math" feature in this forum. (You can click "Show code" to see how I did it.)

(1) I agree with the part where you have:

`cos(0.5880) = 0.83`

However, 

`sin(0.5880) = 0.5547`

So your answer would have been

`sqrt(13)(0.83 + 0.5547i)` ` = 3 + 2i`

(2) Now, the fact our numbers are the wrong way round gives a clue to where the solution went haywire.

The angle representing the complex number `-5+12i` is in the <strong>second quadrant</strong>, so it will be an angle between `pi/2~~1.5708` and `pi~~3.1416`, and `0.5880` is not in this range.

So we need to make use of the <a href="https://staging.intmath.com/trigonometric-functions/6-trigonometry-functions-any-angle.php">Reference angle</a> (about half-way down that page).

Our angle should be

`theta = 0.5(pi - arcsin(12/13))` ` = 0.5xx1.9656` ` = 0.9828`

So now we'll have

`sqrt(13)( cos[0.9828] + i sin [0.9828])` ` = 2+3i`

which is what we were hoping for.

Hope it helps.

X

Thanks so much! Here  I  was worrying that mathematics didn't make sense. You have restored  my faith. 
As I'm sure you realized, I must have  been taking that sin .588 in degrees.

X

re <a href="https://snowrider3d.com">Snow Rider 3D</a>
It's interesting to see your approach to finding the square root of a complex number using polar coordinates. Your calculation for (2+3i)^2 = -5+12i seems correct, and your conversion into polar form with theta = 0.5880 is on point. Have you considered exploring the concept of principal square roots in complex numbers further? It might shed more light on how to determine the correct square root when dealing with complex numbers. Keep up the good work in unraveling the mysteries of complex arithmetic!

X

Re: <a href="https://geometrydash-subzero.io">Geometry Dash</a>

It seems like you're exploring the idea of representing integers as the sum of complex numbers with specific arguments. Murray's point about showing your working is crucial for pinpointing where the challenge lies. Have you tried breaking down how to express those 11 integers using complex numbers with angles of 45°, 135°, and 270°? It might be a fascinating exercise to work through examples to see if patterns emerge based on these arguments. Exploring different approaches step by step could shed light on this intriguing concept further.

X

Nice work showing your steps — you’re very close. The error is in how you find the argument (angle) and in the modulus notation. Corrections and the clean method below.

Mistakes you made

r should be the modulus 
𝑅
=
𝑎
2
+
𝑏
2
R=
a
2
+b
2
	​

, not 
𝑎
2
+
𝑏
2
a
2
+b
2
.

Using 
𝜃
=
1
2
arcsin
⁡
 ⁣
(
𝑏
𝑅
)
θ=
2
1
	​

arcsin(
R
b
	​

) without quadrant correction is unsafe. arcsin returns a principal value in 
[
−
𝜋
/
2
,
𝜋
/
2
]
[−π/2,π/2] and can’t tell whether the point 
(
𝑎
,
𝑏
)
(a,b) is in Q1, Q2, Q3 or Q4. Use atan2(b,a) (or handle quadrants) to get the correct argument 
𝜑
φ.

Correct formula
For 
𝑧
=
𝑎
+
𝑏
𝑖
z=a+bi:

𝑅
=
𝑎
2
+
𝑏
2
R=
a
2
+b
2
	​

 (modulus)

𝜑
=
atan2
⁡
(
𝑏
,
𝑎
)
φ=atan2(b,a) (argument, use atan2 to get the correct quadrant)

Then the square roots are

𝑧
=
±
𝑅
 
(
cos
⁡
(
𝜑
2
)
+
𝑖
sin
⁡
(
𝜑
2
)
)
.
z
	​

=±
R
	​

(cos(
2
φ
	​

)+isin(
2
φ
	​

)).

Apply to your example 
𝑧
=
−
5
+
12
𝑖
z=−5+12i:

𝑅
=
(
−
5
)
2
+
12
2
=
25
+
144
=
169
=
13.
R=
(−5)
2
+12
2
	​

=
25+144
	​

=
169
	​

=13.

𝜑
=
atan2
⁡
(
12
,
−
5
)
=
𝜋
−
arctan
⁡
(
12
/
5
)
≈
1.96559
 rad
.
φ=atan2(12,−5)=π−arctan(12/5)≈1.96559 rad.

𝜑
2
≈
0.982795
 rad
.
2
φ
	​

≈0.982795 rad.

𝑅
=
13
≈
3.60555.
R
	​

=
13
	​

≈3.60555.

cos
⁡
(
𝜑
2
)
≈
0.5547
,
 
sin
⁡
(
𝜑
2
)
≈
0.8321.
cos(
2
φ
	​

)≈0.5547, sin(
2
φ
	​

)≈0.8321.

Multiply: 
𝑅
cos
⁡
(
𝜑
2
)
≈
3.60555
×
0.5547
≈
2.0
,
R
	​

cos(
2
φ
	​

)≈3.60555×0.5547≈2.0,

𝑅
sin
⁡
(
𝜑
2
)
≈
3.60555
×
0.8321
≈
3.0.
R
	​

sin(
2
φ
	​

)≈3.60555×0.8321≈3.0.

So 
−
5
+
12
𝑖
=
±
(
2
+
3
𝑖
)
−5+12i
	​

=±(2+3i). That checks because 
(
2
+
3
𝑖
)
2
=
−
5
+
12
𝑖
(2+3i)
2
=−5+12i.

Bottom line: use 
𝑅
=
𝑎
2
+
𝑏
2
R=
a
2
+b
2
	​

 and 
𝜑
=
atan2
⁡
(
𝑏
,
𝑎
)
φ=atan2(b,a) (not plain arcsin(b/R) without quadrant work). That fixes the “ridiculous” result you got.

If you want, I can show the algebraic method (solve 
(
𝑥
+
𝑖
𝑦
)
2
=
𝑎
+
𝑖
𝑏
(x+iy)
2
=a+ib for 
𝑥
,
𝑦
x,y) or walk through another example. For step-by-step help in similar questions you might consider <p><a href="https://myassignmenthelp.com/take-my-online-class.html">expert online class assistance for students at MyAssignmentHelp.</a></p>