How to find solutions of the equation in Cartesian form?

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Can someone please explain to me how to do question 6? Thanks!

1 Answer

z=1/sqrt2+i/sqrt2
z=-1/sqrt2-i/sqrt2
z=1/sqrt2+i/sqrt2
z=-1/sqrt2-i/sqrt2

Explanation:

Given:

z^2-i=0

z=x+iy

z^2=(x+iy)^2

z^2=x^2+2ixy+(iy)^2

i^2=-1

z^2=(x^2-y^2)+2ixy
substituting for z^2

(x^2-y^2)+2ixy-i=0

(x^2-y^2)+(2xy-1)i=0+0i

Equating the real and imaginary parts

x^2-y^2=0

2xy-1=0

2xy=1

y=1/(2x)

x^2-(1/(2x))^2=0

x^2-1/(4x^2)=0

Let
t=x^2

t-1/(4t)=0

4t^2-1=0

4t^2=1

t^2=1/4

t=+-1/2

t=1/2

t=-1/2

ie

x^2=1/2
x=+-1/sqrt2

x=1/sqrt2

x=-1/sqrt2

x^2=-1/2

x=+-i/sqrt2

x=i/sqrt2

x=-i/sqrt2

Thus,

Substituting for x,

x=1/sqrt2 ,
y=1/(2x)=1/(2xx1/sqrt2)

y=1/sqrt2

x=-1/sqrt2 ,
y=1/(2x)=1/(2xx-1/sqrt2)

y=-1/sqrt2

x=i/sqrt2 ,
y=1/(2x)=1/(2xxi/sqrt2)

y=-i/sqrt2

x=-i/sqrt2 ,
y=1/(2x)=1/(2xx-i/sqrt2)

y=i/sqrt2

We have

(1/sqrt2,1/sqrt2)
(-1/sqrt2,-1/sqrt2)
(i/sqrt2,-i/sqrt2)
(-i/sqrt2,i/sqrt2)
as solutions

z=1/sqrt2+1/sqrt2i
z=-1/sqrt2+-1/sqrt2i
z=i/sqrt2+1/sqrt2
z=-i/sqrt2-1/sqrt2

Rearranging

z=1/sqrt2+i/sqrt2
z=-1/sqrt2-i/sqrt2
z=1/sqrt2+i/sqrt2
z=-1/sqrt2-i/sqrt2