Stuck on expressing a complex number in the form (a+bi)

[Razberryz]

Homework Statement

Express the complex number (−3 +4i)3 in the form a + bi

Homework Equations

z = r(cos(θ) + isin(θ))

The Attempt at a Solution

z = -3 + 4i
z³ = r³(cos(3θ) + isin(3θ))

r = sqrt ((-3)² + 4²)
= 5

θ = arcsin(4/5) = 0.9273

∴ z³ = 5³(cos(3⋅0.9273) + isin(3⋅0.9273))

a = -117
b = 44

Don't see what I did wrong, would appreciate any help.

 

[Fightfish]

z = -3 + 4i
θ = arcsin(4/5) = 0.9273

$\theta$ is in the wrong quadrant. Also, I think it's much easier if you simply expand the original cubic expression instead of converting to polar representation.

 

[Razberryz]

$\theta$ is in the wrong quadrant. Also, I think it's much easier if you simply expand the original cubic expression instead of converting to polar representation.

Thanks for the reply. So you're saying it's not the top left quadrant? Which quadrant is it it then?

 

[Fightfish]

Thanks for the reply. So you're saying it's not the top left quadrant? Which quadrant is it it then?

It is in the top left quadrant - you've got that right. But the value of $\theta$ that you gave is in the first quadrant (i.e. top right quadrant).

 

[Ray Vickson]

Homework Statement

Express the complex number (−3 +4i)3 in the form a + bi

Homework Equations

z = r(cos(θ) + isin(θ))

The Attempt at a Solution

z = -3 + 4i
z³ = r³(cos(3θ) + isin(3θ))

r = sqrt ((-3)² + 4²)
= 5

θ = arcsin(4/5) = 0.9273

Doing the problem that way is bound to lead to numerical roundoff errors, because you cannot find exact decimal value for things like $\arcsin(4/5)$, etc. Why not just go ahead and use direct complex multiplication $(-3+4i)(-3+4i)(-3+4i)$. Guaranteed exact accuracy, no fractions, no decimals, no roundoff errors!

 

[Mark44]

Express the complex number (−3 +4i)3 in the form a + bi

As you wrote it, this is an extremely simple problem
(-3 + 4i)3 = -9 + 12i Done!

It took me some time to realize that the actual problem was to convert (-3 + 4i)³. The simplest way to indicate that 3 is an exponent is to use the caret character ^, as in (-3 + 4i)^3, using notation that comes from the Basic programming language.

 

[Razberryz]

As you wrote it, this is an extremely simple problem
(-3 + 4i)3 = -9 + 12i Done!

It took me some time to realize that the actual problem was to convert (-3 + 4i)³. The simplest way to indicate that 3 is an exponent is to use the caret character ^, as in (-3 + 4i)^3, using notation that comes from the Basic programming language.

Yes I meant, exponent, didn't catch that till later.

 

[lurflurf]

Doing the problem that way is bound to lead to numerical roundoff errors, because you cannot find exact decimal value for things like $\arcsin(4/5)$, etc. Why not just go ahead and use direct complex multiplication $(-3+4i)(-3+4i)(-3+4i)$. Guaranteed exact accuracy, no fractions, no decimals, no roundoff errors!

Roundoff error is clearly not the problem. The long way is only an option because 3 is a small integer if we had
$$(-3+4\imath)^{31763}$$
or
$$(-3+4\imath)^\pi$$
we would not have "Guaranteed exact accuracy, no fractions, no decimals, no roundoff errors!"
I was surprised when I used my sine table in pi/48 increments it was not accurate enough
I needed interpolation to establish arcsin(4/5)~85pi/288
hence
(-3+4i)^3~125(cos(255pi/288)+isin(255pi/288))

Anyway in the original question we need to use quadrant knowledge as mentioned to find that
z^3 = 5^3(-cos(3⋅0.9273) + isin(3⋅0.9273))
or
z^3 = 5^3(cos(3⋅2.2143) + isin(3⋅2.2143))
One way to see this is to use the function arctan2 available on many apps and calculators
arctan2(5,-3)=pi-Arctan(4/3)~2.2143
it is a version of arctan that takes care of quadrants automatically
https://en.wikipedia.org/wiki/Atan2

 

[Ray Vickson]

Roundoff error is clearly not the problem. The long way is only an option because 3 is a small integer if we had
$$(-3+4\imath)^{31763}$$
or
$$(-3+4\imath)^\pi$$
we would not have "Guaranteed exact accuracy, no fractions, no decimals, no roundoff errors!"
I was surprised when I used my sine table in pi/48 increments it was not accurate enough
I needed interpolation to establish arcsin(4/5)~85pi/288
hence
(-3+4i)^3~125(cos(255pi/288)+isin(255pi/288))

Anyway in the original question we need to use quadrant knowledge as mentioned to find that
z^3 = 5^3(-cos(3⋅0.9273) + isin(3⋅0.9273))
or
z^3 = 5^3(cos(3⋅2.2143) + isin(3⋅2.2143))
One way to see this is to use the function arctan2 available on many apps and calculators
arctan2(5,-3)=pi-Arctan(4/3)~2.2143
it is a version of arctan that takes care of quadrants automatically
https://en.wikipedia.org/wiki/Atan2

Of course I realize all that you are saying, but my point was to the OP: was he even aware that straight multiplication could be very effective and that he need not jump instantly to use of the polar representation? Besides, the exact statement of the problem was "Express the complex number (−3 +4i)3 in the form a + bi"; it contains no mention at all of "quadrants". Those become an issue IF one decides to use the polar representation, otherwise not. And, of course, the problem was not anything like $(-3 + 4i)^{31763}$, so tools that would be indispensable for that problem have less relevance for the actual problem he was given.

I take the view that it is not harmful to the OP to be aware of different ways of dealing with a problem. Straight multiplication (repeated squaring, etc.) can be effective in getting $(a+bi)^n$ for moderate integer $n > 1$, perhaps up to about $n = 12$, but of course for some larger values of $n$ the multiplication method becomes increasingly annoying (and increasingly susceptible to making errors), so switching to polar makes sense. And, of course if $n$ is not a positive or negative integer, the polar representation is more-or-less a requirement---not always, but usually.