Solve z=1+i/1+i{1+frac{i/1+i}}

$$z = 1 + \frac { i } { 1 + \frac { i } { 1 + \frac { i } { 1 + i } } }$$

Solve for z

$$z=\frac{4}{3}+\frac{2}{3}i\approx 1.333333333+0.666666667i$$

Steps for Solving Linear Equation

Multiply both numerator and denominator of \frac{i}{1+i} by the complex conjugate of the denominator, 1-i.

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

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

By definition, i^{2} is -1. Calculate the denominator.

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

Multiply i times 1-i.

$$z=1+\frac{i}{1+\frac{i}{1+\frac{i-i^{2}}{2}}}$$

By definition, i^{2} is -1.

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

Do the multiplications in i-\left(-1\right). Reorder the terms.

$$z=1+\frac{i}{1+\frac{i}{1+\frac{1+i}{2}}}$$

Divide 1+i by 2 to get \frac{1}{2}+\frac{1}{2}i.

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

Combine the real and imaginary parts in numbers 1 and \frac{1}{2}+\frac{1}{2}i.

$$z=1+\frac{i}{1+\frac{i}{1+\frac{1}{2}+\frac{1}{2}i}}$$

Add 1 to \frac{1}{2}.

$$z=1+\frac{i}{1+\frac{i}{\frac{3}{2}+\frac{1}{2}i}}$$

Multiply both numerator and denominator of \frac{i}{\frac{3}{2}+\frac{1}{2}i} by the complex conjugate of the denominator, \frac{3}{2}-\frac{1}{2}i.

$$z=1+\frac{i}{1+\frac{i\left(\frac{3}{2}-\frac{1}{2}i\right)}{\left(\frac{3}{2}+\frac{1}{2}i\right)\left(\frac{3}{2}-\frac{1}{2}i\right)}}$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$z=1+\frac{i}{1+\frac{i\left(\frac{3}{2}-\frac{1}{2}i\right)}{\left(\frac{3}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}i^{2}}}$$

By definition, i^{2} is -1. Calculate the denominator.

$$z=1+\frac{i}{1+\frac{i\left(\frac{3}{2}-\frac{1}{2}i\right)}{\frac{5}{2}}}$$

Multiply i times \frac{3}{2}-\frac{1}{2}i.

$$z=1+\frac{i}{1+\frac{\frac{3}{2}i-\frac{1}{2}i^{2}}{\frac{5}{2}}}$$

By definition, i^{2} is -1.

$$z=1+\frac{i}{1+\frac{\frac{3}{2}i-\frac{1}{2}\left(-1\right)}{\frac{5}{2}}}$$

Do the multiplications in \frac{3}{2}i-\frac{1}{2}\left(-1\right). Reorder the terms.

$$z=1+\frac{i}{1+\frac{\frac{1}{2}+\frac{3}{2}i}{\frac{5}{2}}}$$

Divide \frac{1}{2}+\frac{3}{2}i by \frac{5}{2} to get \frac{1}{5}+\frac{3}{5}i.

$$z=1+\frac{i}{1+\left(\frac{1}{5}+\frac{3}{5}i\right)}$$

Combine the real and imaginary parts in numbers 1 and \frac{1}{5}+\frac{3}{5}i.

$$z=1+\frac{i}{1+\frac{1}{5}+\frac{3}{5}i}$$

Add 1 to \frac{1}{5}.

$$z=1+\frac{i}{\frac{6}{5}+\frac{3}{5}i}$$

Multiply both numerator and denominator of \frac{i}{\frac{6}{5}+\frac{3}{5}i} by the complex conjugate of the denominator, \frac{6}{5}-\frac{3}{5}i.

$$z=1+\frac{i\left(\frac{6}{5}-\frac{3}{5}i\right)}{\left(\frac{6}{5}+\frac{3}{5}i\right)\left(\frac{6}{5}-\frac{3}{5}i\right)}$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$z=1+\frac{i\left(\frac{6}{5}-\frac{3}{5}i\right)}{\left(\frac{6}{5}\right)^{2}-\left(\frac{3}{5}\right)^{2}i^{2}}$$

By definition, i^{2} is -1. Calculate the denominator.

$$z=1+\frac{i\left(\frac{6}{5}-\frac{3}{5}i\right)}{\frac{9}{5}}$$

Multiply i times \frac{6}{5}-\frac{3}{5}i.

$$z=1+\frac{\frac{6}{5}i-\frac{3}{5}i^{2}}{\frac{9}{5}}$$

By definition, i^{2} is -1.

$$z=1+\frac{\frac{6}{5}i-\frac{3}{5}\left(-1\right)}{\frac{9}{5}}$$

Do the multiplications in \frac{6}{5}i-\frac{3}{5}\left(-1\right). Reorder the terms.

$$z=1+\frac{\frac{3}{5}+\frac{6}{5}i}{\frac{9}{5}}$$

Divide \frac{3}{5}+\frac{6}{5}i by \frac{9}{5} to get \frac{1}{3}+\frac{2}{3}i.

$$z=1+\left(\frac{1}{3}+\frac{2}{3}i\right)$$

Combine the real and imaginary parts in numbers 1 and \frac{1}{3}+\frac{2}{3}i.

$$z=1+\frac{1}{3}+\frac{2}{3}i$$

Add 1 to \frac{1}{3}.

$$z=\frac{4}{3}+\frac{2}{3}i$$

Assign z

$$z≔\frac{4}{3}+\frac{2}{3}i$$