How to find solutions to equations involving imaginary numbers?

Mistake: i=j?

If so, then

(a+bi)^2=(1+i)/sqrt(2)

=>

sqrt(2)(a^2-b^2)=1------(1)

2sqrt(2)ab=1------------(2)

Putting (2) in (1) get

a^2-1/(8a^2)=1/sqrt(2)=>

8a^4-8a^2/sqrt(2)-1=0=>

a=+/-0.9502081

a=+/-0.4424802i

From (2),

b=0.35355339/a=>

b=+/-2.976639674

b=-/+0.799026464i

Since a, b are real, the answers

are

a=0.9502081, b=2.976639674

a=-0.9502081, b=-2.976639674

I solved this problem using difference of perfect squares.

Therefore, instead of writing (a+bi)^2, I am going to write z^2.

Also, the complex number '(1 + j) /(2)^0.5' is an eyesore, so I am just going to say write 'z_c' where: z_c = (1 + j) /(2)^0.5

(a+bi)^2 = (1 + j) /(2)^0.5

(z)^2 = z_c

(z)^2 - z_c = 0

(z)^2 - [(z_c)^(1/2)]^2 = 0

(z_c)^(1/2) = 1∠(pi/8)

(z)^2 - [1∠(pi/8)]^2 = 0

[ z + 1∠(pi/8) ] * [ z - 1∠(pi/8) ] = 0

[ z + 1∠(pi/8) ] * [ z + 1∠(9*pi/8) ] = 0

[ z_1 + 1∠(pi/8) ] * [ z_2 + 1∠(9*pi/8) ] = 0

Therefore, the two solutions are shown below

z_1 = -1∠(pi/8) = 1∠(9*pi/8)

z_2 = - 1∠(9*pi/8 = 1∠(pi/8)

What are you solving for? Remember that both i and j are used to denote √-1

Let us use j rather than i

Presentation is open to doubt so having to take a guess at :-

[ a + j b ]