Solve z=frac{1+isqrt{3}}{1+i}

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

Solve for z

$$z=\sqrt{3}\left(\frac{1}{2}+\frac{1}{2}i\right)+\left(\frac{1}{2}-\frac{1}{2}i\right)\approx 1.366025404+0.366025404i$$

Steps for Solving Linear Equation

Divide each term of 1+i\sqrt{3} by 1+i to get \frac{1}{1+i}+\frac{i\sqrt{3}}{1+i}.

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

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

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

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

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

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

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

Multiply 1 and 1-i to get 1-i.

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

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

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

Divide i\sqrt{3} by 1+i to get \left(\frac{1}{2}+\frac{1}{2}i\right)\sqrt{3}.

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

Assign z

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