Question

How do i solve for z?
(z+1)^2 = 3 + 4i

Answer 1

write z = x+iy

Answer 2

\((x+iy+1)^2 = ([x+1] + iy )^2 =.....\)

use \( (a+b)^2 = a^2+2ab+b^2\) formula

Answer 3

Oh.. let me try that out

Answer 4

What I have so far now is...
\[(x+1)^{2} +2iy(x+1) - y ^{2} = 3 +4i\]

Answer 5

yes, correct.
go on

Answer 6

Thanks. I will try and expand and see what happens

Answer 7

Hmm.. not sure if I went the wrong way here...
\[x ^{2} +2x - y ^{2} +1 + i(2xy +2y) = 3 + 4i\]
Not sure how to go about putting in z = x + iy

Answer 8

actually, i can think of a better way...
why not let u = z+1

then u^2 = 3+4i

Answer 9

it would be much easier to find u
then we can always re substitute u = z+1

Answer 10

and letting u = x+iy

Answer 11

I see. And I believe I can use some equations I have here
letting a = 3 and b =4
Thanks! I am going to try that approach

Answer 12

Ok, I got close it seems..
\[z = 1 +\frac{ 1 }{ \sqrt{2} }i
and z = -3 - \frac{ 1 }{ \sqrt{2} }i\]

The solutions are suppose to be z = 1+i and z =-3 -i

Answer 13

x^2 - y^2 = 3
2xy = 4

you got these ?

Answer 14

No..
I used some formulas that used alpha and beta..
Probably should have worked to but I see you replaced
I see how you got those though...
Let me try 1 more time!!
Thanks

Answer 15

I have an idea, but you guys would probably curse me for even suggesting it ^_^

...polar form, anyone?

Answer 16

haha, why not! if it reduces any efforts :P

Answer 17

\[\Large 3+4\color{blue}i = 5\left[\cos\theta + \color{blue}i\sin\theta\right]\]

Answer 18

So that... to get the solution of
\[\Large (z+1)^2 = 3+4\color{blue}i = 5\left[\cos\theta + \color{blue}i\sin\theta\right]\]

Answer 19

I going to try both as this question is before the section on polar form but I need to know both

Answer 20

We just have to get the two square roots of the complex-number to the right....

Answer 21

Before polar form, huh?
Should I proceed anyway? ^_^

Answer 22

yes, we'll discuss both methods here.

Answer 23

Yes. Proceed haha

Answer 24

All right. Let me just get this off the table: ^_^
\[\Large z = r\left[\cos\theta + \color{blue}i\sin\theta\right]\]\[\Large z^\color{green}p=r^\color{green}p\left[\cos \color{green}p\theta + \color{blue}i\sin \color{green}p\theta \right]\]

This is DeMoivre's theorem... it works for fractional powers too... (like square roots).... sort of....
So...

Answer 25

\[\Large (z+1)^2 = 5\left[\cos\theta + \color{blue}i\sin\theta\right]\]
Get the square root of both sides, we get:
\[\Large z+1 =\sqrt 5\left[\cos\frac{\theta}2 + \color{blue}i\sin\frac{\theta}2\right]\]

Since square root is the same as raising to a power 1/2.

Answer 26

Relating to first method...Using
x^2 - y^2 = 3
2xy = 4
I got.... let y = 2/x and plugged to top equation and got for x values...
x = +-2 +-i i got the corresponding y's although not sure what to do now

Answer 27

NOTE : in u = x+iy
x and y are real numbers!

Answer 28

The polar form is looking good.
I have seen those equations but needed to check how theta got divided by 2 but I see now, like you said

Answer 29

We don't (actually) know what \(\theta\) is... (ugly, whether in degrees or radians) But we DO know that
\[\Large \cos \theta = \frac35\] and this is actually all we need ^_^

Remembering half-angle identities...
\[\Large \cos\frac\theta 2=\sqrt{\frac{1+\cos\theta}{2}}=\sqrt{\frac4{5}}\]after substituting.

similarly...\[\Large \sin\frac\theta 2=\sqrt{\frac{1-\cos\theta}{2}}=\sqrt{\frac1{5}}\]

Answer 30

Alright I will use the real numbers for x and y

Answer 31

Back to the polar form, we get:
\[\Large z+1 =\sqrt 5\left[\cos\frac{\theta}2 + \color{blue}i\sin\frac{\theta}2\right]\]\[\Large z+1 =\sqrt 5\left[\sqrt{\frac45}+ \color{blue}i\sqrt{\frac15}\right]=2+\color{blue}i\]
And the rest is history ^_^

Answer 32

Nice! And I believe I can find the other solution

Answer 33

As for the other root, we could go back to here:
\[\Large (z+1)^2 = 5\left[\cos\theta + \color{blue}i\sin\theta\right]\]
And just note that
\[\Large (z+1)^2 = 5\left[\cos(\theta+2\pi) + \color{blue}i\sin(\theta+2\pi)\right]\]

And again, get the square root...
\[\Large z+1 = \sqrt5\left[\cos\frac{(\theta+2\pi)}2 + \color{blue}i\sin\frac{(\theta+2\pi)}2\right]\]
\[\Large z+1 =\sqrt 5\left[\cos\left(\frac{\theta}2+\pi\right) + \color{blue}i\sin\left(\frac{\theta}2+\pi\right) \right]\]

Answer 34

Oh sorry, feel free to proceed, then ^_^

Answer 35

Thank you both!!
I just finished using the other method and just to be safe..
We using half-angle identities for this solution as well?

Answer 36

Of course, and of course, the fact that...
\[\Large \cos(\alpha+\pi) = -\cos(\alpha)\\\Large \sin(\alpha+\pi) = -\sin(\alpha)\]
good hunting ^_^

Note that I didn't include the +/- bit in the half-angle identity because the argument \(\theta\) is in the first quadrant (due to the real and imaginary parts of 3+4i both being positive)

Answer 37

Great! All I do is change the sign thanks to pie

Answer 38

Yup. And you get the other answer, am I right? ^_^

Answer 39

Yep!! I just did.
Thanks a bunch!