+0

# Ok. I think this might be calculus. real hard.

+15
284
7
+1144

All help is greatly appreciated for this problem.

Let $$z = a + bi,$$ where a and b are positive real numbers. If

$$z^3 + |z|^2 + z = 0,$$
then enter the ordered pair $$(a,b).$$

Jul 8, 2022

#1
0

(a,b) = (1,sqrt(3)).

Jul 8, 2022
#2
+1144
+11

No explanation, and it's wrong.

nerdiest  Jul 8, 2022
#3
+118594
+2

Hi Nerdiest.

I just worked out each bit seperately and then put it altogether.

I have not presented it becasue it is a bit long and I no doubt made careless errors

But I will walk you through it

$$z=a+bi\\ z^3=a^3+3a^2bi+3a(bi)^2+(bi)^3=etc\\ |z|^2=a^2+b^2$$

The real part is 0 and the imaginary part is 0

Solve simultaneously.

Jul 9, 2022
#4
+1

Hi.

Given:                                                    $$z=a+bi$$

So:                           $$z^3=(a+bi)^3=a^3+3a^2bi-3ab^2-b^3i$$

Given:                                   $$z^3+\left | z\right |^2+z=0$$

Substitute $$z,z^3$$, and we know: $$\left |z \right |^2=\sqrt{a^2+b^2}^2=a^2+b^2$$

$$a^3+3a^2bi-3ab^2-b^3i+(a^2+b^2)+a+bi=0$$

We equate real parts = 0 and imaginary parts = 0             (As the right-hand side is zero, this means both the real and imaginary parts are zero.) (Real parts without the "i" and imaginary parts with the "i").

So: Set $$Re=0:$$  $$a^3-3ab^2+a^2+b^2+a=0$$       (1)

Set $$Im=0:$$   $$3a^2b-b^3+b=0$$                          (2)

Given b is not equal to zero so, divide (2) by b:

$$3a^2-b^2+1=0$$

Let's make b the subject of the equation: (That is, solve for b.)

$$b=\sqrt{3a^2+1}$$

Hence, substituting b in (1) to get:

$$a^3-3a(3a^2+1)+a^2+3a^2+1+a=0$$   (Expand.)

$$a^3-9a^3-3a+a^2+3a^2+1+a=0$$        (Simplify.)

$$-8a^3-2a+4a^2+1=0$$                               (Multiply by -1)

$$8a^3-4a^2+2a-1=0$$

We can factor this by grouping:

$$4a^2(2a-1)+(2a-1)=0$$

$$(2a-1)(4a^2+1)=0$$

We know that a is a positive real number.

So we only take:  $$a=\frac{1}{2}$$

Hence, $$b=\sqrt{3(0.5)^2+1}=\sqrt{\frac{7}{4}}=\frac{\sqrt{7}}{2}$$

Therefore, $$(\frac{1}{2},\frac{{\sqrt{7}}}{2})$$ is the desired answer.

I hope this helps :)!

Jul 9, 2022
edited by Guest  Jul 9, 2022
#5
+118594
+3

Incidentally, this has nothing to do with calculus.

Jul 9, 2022
#6
+1144
+10

THANK YOU VERY MUCH MELODY AND GUEST, for I got the first insanely hard, than easy understanding question RIGHT!

THANK YOU

THANK YOU

THANK YOU

nerdiest  Jul 9, 2022
#7
+118594
+1

You are very welcome Nerdiest :)

Melody  Jul 10, 2022