Solve 4/3+sqrt{-1}

$$\frac{ 4 }{ 3+ \sqrt{ -1 } }$$

Evaluate (complex solution)

$$\frac{6}{5}-\frac{2}{5}i=1.2-0.4i$$

Solution Steps

Calculate the square root of -1 and get i.

$$\frac{4}{3+i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, 3-i.

$$\frac{4\left(3-i\right)}{\left(3+i\right)\left(3-i\right)}$$

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

$$\frac{4\left(3-i\right)}{3^{2}-i^{2}}$$

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

$$\frac{4\left(3-i\right)}{10}$$

Multiply 4 times 3-i.

$$\frac{4\times 3+4\left(-i\right)}{10}$$

Do the multiplications in 4\times 3+4\left(-i\right).

$$\frac{12-4i}{10}$$

Divide 12-4i by 10 to get \frac{6}{5}-\frac{2}{5}i.

$$\frac{6}{5}-\frac{2}{5}i$$

Real Part (complex solution)

$$\frac{6}{5} = 1\frac{1}{5} = 1.2$$

Solution Steps

Calculate the square root of -1 and get i.

$$Re(\frac{4}{3+i})$$

Multiply both numerator and denominator of \frac{4}{3+i} by the complex conjugate of the denominator, 3-i.

$$Re(\frac{4\left(3-i\right)}{\left(3+i\right)\left(3-i\right)})$$

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

$$Re(\frac{4\left(3-i\right)}{3^{2}-i^{2}})$$

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

$$Re(\frac{4\left(3-i\right)}{10})$$

Multiply 4 times 3-i.

$$Re(\frac{4\times 3+4\left(-i\right)}{10})$$

Do the multiplications in 4\times 3+4\left(-i\right).

$$Re(\frac{12-4i}{10})$$

Divide 12-4i by 10 to get \frac{6}{5}-\frac{2}{5}i.

$$Re(\frac{6}{5}-\frac{2}{5}i)$$

The real part of \frac{6}{5}-\frac{2}{5}i is \frac{6}{5}.

$$\frac{6}{5}$$

Evaluate

$$\text{Indeterminate}$$

Solution Steps

Rationalize the denominator of \frac{4}{3+\sqrt{-1}} by multiplying numerator and denominator by 3-\sqrt{-1}.

$$\frac{4\left(3-\sqrt{-1}\right)}{\left(3+\sqrt{-1}\right)\left(3-\sqrt{-1}\right)}$$

Consider \left(3+\sqrt{-1}\right)\left(3-\sqrt{-1}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{4\left(3-\sqrt{-1}\right)}{3^{2}-\left(\sqrt{-1}\right)^{2}}$$

Square 3. Square \sqrt{-1}.

$$\frac{4\left(3-\sqrt{-1}\right)}{9+1}$$

Subtract -1 from 9 to get 10.

$$\frac{4\left(3-\sqrt{-1}\right)}{10}$$

Divide 4\left(3-\sqrt{-1}\right) by 10 to get \frac{2}{5}\left(3-\sqrt{-1}\right).

$$\frac{2}{5}\left(3-\sqrt{-1}\right)$$

Use the distributive property to multiply \frac{2}{5} by 3-\sqrt{-1}.

$$\frac{2}{5}\times 3+\frac{2}{5}\left(-1\right)\sqrt{-1}$$

Express \frac{2}{5}\times 3 as a single fraction.

$$\frac{2\times 3}{5}+\frac{2}{5}\left(-1\right)\sqrt{-1}$$

Multiply 2 and 3 to get 6.

$$\frac{6}{5}+\frac{2}{5}\left(-1\right)\sqrt{-1}$$

Multiply \frac{2}{5} and -1 to get -\frac{2}{5}.

$$\frac{6}{5}-\frac{2}{5}\sqrt{-1}$$