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# How can I find the argument (angle) of -81i?

This is a portion from the problem: Find Solution strictly between 90 degrees and 180 degrees for z^(2) = -81i. I know the modulus is 81, but I don't know how to find the argument when only 81i is given.

Thanks.

### 7 Answers

- PinkgreenLv 71 month ago
z^2=-81i=>

z^2=81(-i)=>

z^2=81e^[(2kpi-pi/2)i]=>

z=9e^[(kpi-pi/4)i]

z0=9e^[(-pi/4)i], k=0

z1=9e^[(3pi/4)i], k=1

=>

the argument in (pi/2, pi) is 3pi/4.

- Wayne DeguManLv 71 month ago
-81i = 81[cos(270° + 360n°) + isin(270° + 360n°)]

so, √-81i = 9[cos(270° + 360n°) + isin(270° + 360n°)] ¹/²

i.e. √-81i = 9[cos(135° + 180n°) + isin(135° + 180n°)]

With n = 0 we have:

9[cos135° + isin135°] => 9[-√2/2 + i√2/2]

or, (9√2/2)[-1 + i]

With n = 1 we have:

9[cos315° + isin315°] => 9[√2/2 - i√2/2]

or, (9√2/2)[1 - i]

Hence, ±(9√2/2)[1 - i]

Note, squaring will give -81i

:)>

- DixonLv 71 month ago
Doesn't anyone use polar coordinates???

(A ∠a) x (B ∠b) = (A x B) ∠(a + b)

So the primary solution of √(A ∠a) is just √A ∠a/2

And in general, when taking the nth root of a complex number there are n solutions. These solutions are all the same length (modulus) and are spaced out regularly at angles of 360°/n like spokes of a wheel, centered around the primary solution.

So there will be two solutions which will be in opposite directions, and in your head you can just look at z² = -81i and go

z² = -81i

z² = 81 ∠270°

z = 9 ∠(270°/2) and another solution at ± 180° from that

ie

z = 9 ∠135 , 9 ∠-45°

- ted sLv 71 month ago
- 81 i = 81 ( - i ) = 81 ( cos 3π / 2 + i sin 3π / 2 ) ===> no angle in [ 90° , 180° ]...but you want z² = - 81 i = 81 ( cos Θ+ i sin Θ , where Θ = 270 + n 360 which implies z = 9 ( cos Θ/2 + i sin Θ/2 ) , Θ / 2 = 135 + n 180 ....135° is in [ 90° , 180° ]

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- husoskiLv 71 month ago
Pure imaginary values are on the "y-axis", so positive multiples of i all have angle pi/2 and negative multiples of i all have have angle 3*pi/2 or -pi/2.

You kinda need to just know that since the arctan(y/x) formula blows up there.

I tend to use arccos(x/r) instead in computer programs for that reason, since I usually need to know both r and θ, (The sign of θ is negative, -arccos(x/r), when y < 0.)

- 1 month ago
z^2 = -81i

z^2 = 81 * i^(3 + 4k)

z = 9 * i^(1.5 + 2k)

z = 9 * i^(1.5) , 9 * i^(3.5)

i^(1.5) =>

(0 + i)^(1.5) =>

(cos(pi/2) + i * sin(pi/2))^(1.5) =>

cos(1.5 * pi/2) + i * sin(1.5 * pi/2) =>

cos(3pi/4) + i * sin(3pi/4)

i^(3.5) =>

(0 + i)^(3.5) =>

(cos(pi/2) + i * sin(pi/2))^(7/2) =>

cos(7pi/4) + i * sin(7pi/4)

3pi/4 => 135 degrees

7pi/4 => 315 degrees

90 < t < 180

t = 135 degrees

The modulus is 9

- Jeff AaronLv 71 month ago
Let z = a + bi, where a and b are real and i^2 = -1, so we have:

(a + bi)^2 = -81i

a^2 + 2abi + b^2i^2 + 81i = 0

a^2 - b^2 + 2abi + 81i = 0

Therefore:

a^2 - b^2 = 0, so a^2 = b^2, so a = +/- b

2ab + 81 = 0, so 2ab = -81, so ab = -81/2 = -40.5

So +/- b * b = -40.5

b^2 = +/- 40.5

Since b is real, b^2 can't be negative, so we have:

b^2 = 40.5

b = +/- sqrt(40.5)

If b = sqrt(40.5) then a = -sqrt(40.5), and vice versa.

So z = sqrt(40.5) - sqrt(-40.5) or z = -sqrt(40.5) + sqrt(-40.5)

Note that sqrt(40.5) =~ 6.3639610306789277196075992589436

And sqrt(-40.5) =~ 6.3639610306789277196075992589436 i