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/sqrt2z=12+i2
z=-1/sqrt2-i/sqrt2z=12i2
z=1/sqrt2+i/sqrt2z=12+i2
z=-1/sqrt2-i/sqrt2z=12i2

Explanation:

Given:

z^2-i=0z2i=0

z=x+iyz=x+iy

z^2=(x+iy)^2z2=(x+iy)2

z^2=x^2+2ixy+(iy)^2z2=x2+2ixy+(iy)2

i^2=-1i2=1

z^2=(x^2-y^2)+2ixyz2=(x2y2)+2ixy
substituting for z^2

(x^2-y^2)+2ixy-i=0(x2y2)+2ixyi=0

(x^2-y^2)+(2xy-1)i=0+0i(x2y2)+(2xy1)i=0+0i

Equating the real and imaginary parts

x^2-y^2=0x2y2=0

2xy-1=02xy1=0

2xy=12xy=1

y=1/(2x)y=12x

x^2-(1/(2x))^2=0x2(12x)2=0

x^2-1/(4x^2)=0x214x2=0

Let
t=x^2t=x2

t-1/(4t)=0t14t=0

4t^2-1=04t21=0

4t^2=14t2=1

t^2=1/4t2=14

t=+-1/2t=±12

t=1/2t=12

t=-1/2t=12

ie

x^2=1/2x2=12
x=+-1/sqrt2x=±12

x=1/sqrt2x=12

x=-1/sqrt2x=12

x^2=-1/2x2=12

x=+-i/sqrt2x=±i2

x=i/sqrt2x=i2

x=-i/sqrt2x=i2

Thus,

Substituting for x,

x=1/sqrt2 , x=12,
y=1/(2x)=1/(2xx1/sqrt2)y=12x=12×12

y=1/sqrt2y=12

x=-1/sqrt2 , x=12,
y=1/(2x)=1/(2xx-1/sqrt2)y=12x=12×12

y=-1/sqrt2y=12

x=i/sqrt2 , x=i2,
y=1/(2x)=1/(2xxi/sqrt2)y=12x=12×i2

y=-i/sqrt2y=i2

x=-i/sqrt2 , x=i2,
y=1/(2x)=1/(2xx-i/sqrt2)y=12x=12×i2

y=i/sqrt2y=i2

We have

(1/sqrt2,1/sqrt2)(12,12)
(-1/sqrt2,-1/sqrt2)(12,12)
(i/sqrt2,-i/sqrt2)(i2,i2)
(-i/sqrt2,i/sqrt2)(i2,i2)
as solutions

z=1/sqrt2+1/sqrt2iz=12+12i
z=-1/sqrt2+-1/sqrt2iz=12±12i
z=i/sqrt2+1/sqrt2z=i2+12
z=-i/sqrt2-1/sqrt2z=i212

Rearranging

z=1/sqrt2+i/sqrt2z=12+i2
z=-1/sqrt2-i/sqrt2z=12i2
z=1/sqrt2+i/sqrt2z=12+i2
z=-1/sqrt2-i/sqrt2z=12i2