Solve z=i(1-ji)/1+i

$$z = \frac { i ( 1 - j i ) } { 1 + i }$$

Solve for j

$$j=\left(1+i\right)z-i$$

Steps for Solving Linear Equation

Divide i\left(1-ji\right) by 1+i to get \left(\frac{1}{2}+\frac{1}{2}i\right)\left(1-ji\right).

$$z=\left(\frac{1}{2}+\frac{1}{2}i\right)\left(1-ji\right)$$

Swap sides so that all variable terms are on the left hand side.

$$\left(\frac{1}{2}+\frac{1}{2}i\right)\left(1-ji\right)=z$$

Multiply -1 and i to get -i.

$$\left(\frac{1}{2}+\frac{1}{2}i\right)\left(1-ij\right)=z$$

Use the distributive property to multiply \frac{1}{2}+\frac{1}{2}i by 1-ij.

$$\frac{1}{2}+\frac{1}{2}i+\left(\frac{1}{2}-\frac{1}{2}i\right)j=z$$

Subtract \frac{1}{2}+\frac{1}{2}i from both sides.

$$\left(\frac{1}{2}-\frac{1}{2}i\right)j=z-\left(\frac{1}{2}+\frac{1}{2}i\right)$$

The equation is in standard form.

$$\left(\frac{1}{2}-\frac{1}{2}i\right)j=z+\left(-\frac{1}{2}-\frac{1}{2}i\right)$$

Divide both sides by \frac{1}{2}-\frac{1}{2}i.

$$\frac{\left(\frac{1}{2}-\frac{1}{2}i\right)j}{\frac{1}{2}-\frac{1}{2}i}=\frac{z+\left(-\frac{1}{2}-\frac{1}{2}i\right)}{\frac{1}{2}-\frac{1}{2}i}$$

Dividing by \frac{1}{2}-\frac{1}{2}i undoes the multiplication by \frac{1}{2}-\frac{1}{2}i.

$$j=\frac{z+\left(-\frac{1}{2}-\frac{1}{2}i\right)}{\frac{1}{2}-\frac{1}{2}i}$$

Divide z+\left(-\frac{1}{2}-\frac{1}{2}i\right) by \frac{1}{2}-\frac{1}{2}i.

$$j=\left(1+i\right)z-i$$

Solve for z

$$z=\left(\frac{1}{2}-\frac{1}{2}i\right)j+\left(\frac{1}{2}+\frac{1}{2}i\right)$$

Solution Steps

Divide i\left(1-ji\right) by 1+i to get \left(\frac{1}{2}+\frac{1}{2}i\right)\left(1-ji\right).

$$z=\left(\frac{1}{2}+\frac{1}{2}i\right)\left(1-ji\right)$$

Multiply -1 and i to get -i.

$$z=\left(\frac{1}{2}+\frac{1}{2}i\right)\left(1-ij\right)$$

Use the distributive property to multiply \frac{1}{2}+\frac{1}{2}i by 1-ij.

$$z=\frac{1}{2}+\frac{1}{2}i+\left(\frac{1}{2}-\frac{1}{2}i\right)j$$