[x-(1*i)][x-(1-i)]=

Simple and best practice solution for [x-(1*i)][x-(1-i)]= equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework.

If it's not what You are looking for type in the equation solver your own equation and let us solve it.

Solution for [x-(1*i)][x-(1-i)]= equation:


Simplifying
[x + -1(1i)][x + -1(1 + -1i)] = 0

Remove parenthesis around (1i)
[x + -1 * 1i][x + -1(1 + -1i)] = 0

Multiply -1 * 1
[x + -1i][x + -1(1 + -1i)] = 0

Reorder the terms:
[-1i + x][x + -1(1 + -1i)] = 0
[-1i + x][x + (1 * -1 + -1i * -1)] = 0
[-1i + x][x + (-1 + 1i)] = 0

Reorder the terms:
[-1i + x][-1 + 1i + x] = 0

Multiply [-1i + x] * [-1 + 1i + x]
[-1i * [-1 + 1i + x] + x[-1 + 1i + x]] = 0
[[-1 * -1i + 1i * -1i + x * -1i] + x[-1 + 1i + x]] = 0

Reorder the terms:
[[1i + -1ix + -1i2] + x[-1 + 1i + x]] = 0
[[1i + -1ix + -1i2] + x[-1 + 1i + x]] = 0
[1i + -1ix + -1i2 + [-1 * x + 1i * x + x * x]] = 0

Reorder the terms:
[1i + -1ix + -1i2 + [1ix + -1x + x2]] = 0
[1i + -1ix + -1i2 + [1ix + -1x + x2]] = 0

Reorder the terms:
[1i + -1ix + 1ix + -1i2 + -1x + x2] = 0

Combine like terms: -1ix + 1ix = 0
[1i + 0 + -1i2 + -1x + x2] = 0
[1i + -1i2 + -1x + x2] = 0

Solving
1i + -1i2 + -1x + x2 = 0

Solving for variable 'i'.

The solution to this equation could not be determined.

See similar equations:

| 2+0.14x=3 | | xyz=4(x+y+z) | | 2(s^2+25)=14 | | x(3x^2+x-2)-2(x^3-x)= | | 3/4-1/5x=6x-5 | | (-3x^5)^2/(-2x^3)^-3 | | 6y^2=100 | | 1/6x-x/4=2 | | S=2rh+2r2 | | 190x+0.25=342.5 | | Logx+log(x-4)=1 | | m^2+2=45 | | (Y-1)(y+1)=48 | | 8x=9x-72 | | 5y+3=-6y+4*(1+3y) | | 4(5z-1)-4(z+4)=5(z+1) | | 2x-4(x-2)=-3+2x-1 | | (3xy^-2)/(12x^-2y^4) | | -15=14x^2-31x | | -1(2+3)=-4+3 | | 0.5x-9=4 | | x^3-x^2=42x | | 4+0.14x=6 | | 8x+3y+67=73d | | x^2-11+30= | | 21+29m=35+22m | | x+x+x=-12 | | 7+21m=35+14m | | (4x+20)+(5x+2)=180 | | -8(y+5)+-29=43 | | 9/4x*4/6 | | 3x-2x=120 |

Equations solver categories