Solve the equation

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http://img87.imageshack.us/img87/8819/eqta4.gif

Please explain how to solve this equation.

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I also have problem with this equation.
http://img221.imageshack.us/img221/5183/gifsk9.gif

Please help me.

[Shevek]

Second one, set z+i=w, then

w^2=1-i

One way to find the two values is to note that if w=r cis v (cis v = cos v + isin v) then

|w|=|r|, and since

|w^2|=|1-i|=r^2=sqrt(2) (the length), then

r=sqrt(sqrt(2)).

Also,

w^2=r^2 cis 2v (de Moivre's formula), so

2v_1=7pi/4 (check the argument/angle of (1-i) in an Im-Re-coordinate system)
2v_2=7pi/4+2pi. (Since it's w to the power of 2, we know that there'll be two solutions, we could keep adding +2pi, 2v_3=7pi/4+4pi et c, but it wouldn't give any new solutions).


and

v_1=7pi/8
v_2=15pi/8.

That gives

z=w-i

z_1=-i+sqrt(sqrt(2))cis 7pi/8
z_1=-i+sqrt(sqrt(2))cis 15pi/8.


Is it possible to understand anything at all of that solution? :oh

[Shevek]

First one.

z(1+i)^2=z*-1-i

2iz=z*-1-i

Write z=a+bi and z*=a-bi, then

2ia-2b=a-1-(b+1)i.

The imaginary part of VL must be equal to the imaginary part of HL. The same goes for the real part, so

-2b=a-1 <--> -4b+2=2a
2a=-b-1

and subtraction gives

0=-3b+3 <--> b=1, so

a=-2b+1=-1.


Answer: z=-1+i

[∅]

This one then (solve the equation)

z^2 + z* + 1 = 0

[Shevek]

I think your teacher wants you to solve them... =P

z=a+bi
z*=a-bi

z^2+z*+1=0
(a+bi)^2+a-bi+1=0
a^2+a-b^2+1+2abi-bi=0

Imaginary parts equal gives b(2a-1)=0, that is b=0 or a=1/2.

Real parts equal and b=0 gives

a^2+a+1=0
(a+1/2)+3/4=0, which has no real solutions.

Real parts equal and a=1/2 gives

1/4+1/2-b^2+1=0
b^2=7/4
b_1=sqrt(7)/2
b_2=-sqrt(7)/2.

So

z_1=(1+isqrt(7))/2
z_2=(1-isqrt(7))/2.

Checking z_1: z^2+z*+1=1/4-7/4+isqrt(7)+1/2-isqrt(7)+1=0
z_2: 1/4-7/4-isqrt(7)+1/2+isqrt(7)+1=0.


Answer: z_1=(1+isqrt(7))/2, z_2=(1-isqrt(7))/2. Note that z_1=z_2*.