# Math Help - complex equation

1. ## complex equation

Question is

(z+i)^3 + 8i = 0

(z+i)^3 = 8i
(z+i)^3 = 2^3 e^(3 pie i / 2 + 2pie i k) where k is any integer

How did we get this step. Can u explain in detail how we got to this step?
(z+i)^3 = 2^3 e^(3 pie i / 2 + 2pie i k) where k is any integer

2. Originally Posted by usman206
Question is

(z+i)^3 + 8i = 0

(z+i)^3 = 8i
(z+i)^3 = 2^3 e^(3 pie i / 2 + 2pie i k) where k is any integer

How did we get this step. Can u explain in detail how we got to this step?
(z+i)^3 = 2^3 e^(3 pie i / 2 + 2pie i k) where k is any integer
Hi, I am NO EXPERT, but it looks like this to me;

(z+i)^3 + 8i = 0 (If +8i on left of equals is positive, then by moving it to the right hand side, should it not be);
(z+i)^3 = -8i

Carry out operations in brackets first.z+i x z+i x z+i = 2^3 = 2x2x2 e^ 3pie = (9.42 i / 8.28 i k)

All it seems to be doing to me is expanding the terms, so by example 8i = 2x2x2 = 8i, just carry on through to the end.

3. Originally Posted by usman206
Question is

(z+i)^3 + 8i = 0

(z+i)^3 = 8i
(z+i)^3 = 2^3 e^(3 pie i / 2 + 2pie i k) where k is any integer

How did we get this step. Can u explain in detail how we got to this step?
(z+i)^3 = 2^3 e^(3 pie i / 2 + 2pie i k) where k is any integer
Do you know how to find the polar form of a complex number such as -8i?

In this case $2 \pi n$ is also added to the argument of -8i so that all three solutions can be got (otherwise you only get one solution).

Originally Posted by David Green
Hi, I am NO EXPERT, but it looks like this to me;

(z+i)^3 + 8i = 0 (If +8i on left of equals is positive, then by moving it to the right hand side, should it not be);
(z+i)^3 = -8i

Mr F says: The above part is helpful.

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Mr F says: The below part is not helpful (except for the bit that suggests that 8 = 2^3)

Carry out operations in brackets first.z+i x z+i x z+i = 2^3 = 2x2x2 e^ 3pie = (9.42 i / 8.28 i k)

All it seems to be doing to me is expanding the terms, so by example 8i = 2x2x2 = 8i, just carry on through to the end.
Knowledge of getting the polar form of -8i is required.

4. Originally Posted by usman206
Question is

(z+i)^3 + 8i = 0

(z+i)^3 = 8i
(z+i)^3 = 2^3 e^(3 pie i / 2 + 2pie i k) where k is any integer

How did we get this step. Can u explain in detail how we got to this step?
(z+i)^3 = 2^3 e^(3 pie i / 2 + 2pie i k) where k is any integer
By putting 8i into polar form.

To put a complex number into polar form we get:

$a+ib = re^{i\theta}$

r is the magnitude of the complex number, given by $|a+ib| = \sqrt{a^2+b^2}$

And $\theta$ is the angle of the complex number given by $\theta = \arctan\bigg(\frac{b}{a}\bigg)$. And remember that if you rotates a complex number by 2pi radians, you arrive at the same angle, so $\theta = \arctan\bigg(\frac{b}{a}\bigg) \pm 2k\pi$

In your case $a = 0, b = -8$, and hence: $r = |8i| = \sqrt{(-8)^2} = 8 = 2^3$

We don't actually need to use the arctan rule in this case, since it's easy to see the angle if you draw it. If you draw the complex number on an Argand diagram you'll see that it is parallel to the imaginary axis, and points downwards. That is 270 degrees (or $\frac{3\pi}{2}$ radians) from the positive real axis. Hence $\theta = \frac{3\pi}{2} \pm 2k\pi$

5. STILL DONT GET IT

CAN WE USE THIS FORMULA TO GET THETA

cos theta = x / z modulus
and
sin theta = y/ z modulus

Can u convert -8i into polar form plz

6. Originally Posted by usman206
Can u convert -8i into polar form plz . plz explain in detail how u get tehta or argument

i normally finda theta bu sing this formula

cos theta = x / z modulus
and sin theta - y / z modulus
Think of the complex plane:
x-axis (horizontal axis) = real axis
y-axis (vertical axis) = imaginary axis

Now think of a complex number $z = a + ib$ as a set of coordinates on this plane $\Rightarrow (a,b)$

Now for $z = -8i$, this corresponds to no real part to your complex number, thus your "x-coordinate" would be zero. Thus your complex number would translate to coordinates $(0,-8)$.

From this picture, you should be able to see visually what the argument of z is.

7. I am no expert as I mentioned previously, my understanding of your subject at the moment is limited, therefore i suggest you read Stroud Engineering maths, complex numbers in polar form. Having briefly read it the understanding is not to difficult, give it a go.

regards

David

8. Originally Posted by usman206
Can u convert -8i into polar form plz
If you need to ask this question then you shouldn't be attempting the question you posted. You need to go back and urgently review how to get the polar form of a complex number.

If you draw -8i on an Argand diagram it should be completely obvious what the polar form is if you have any understanding of the polar form: $8 cis \left( - \frac{\pi}{2}\right)$.