Solve 240/25+25sqrt{3i+10+sqrt{300}i}

$$\frac { 240 } { 25 + 25 \sqrt { 3 } i + 10 + \sqrt { 300 } i }$$

Evaluate

$$\frac{48}{7\left(1+\sqrt{3}i\right)}\approx 1.714285714-2.969229956i$$

Solution Steps

Add 25 and 10 to get 35.

$$\frac{240}{35+25i\sqrt{3}+i\sqrt{300}}$$

Factor 300=10^{2}\times 3. Rewrite the square root of the product \sqrt{10^{2}\times 3} as the product of square roots \sqrt{10^{2}}\sqrt{3}. Take the square root of 10^{2}.

$$\frac{240}{35+25i\sqrt{3}+i\times 10\sqrt{3}}$$

Combine 25i\sqrt{3} and 10i\sqrt{3} to get 35i\sqrt{3}.

$$\frac{240}{35+35i\sqrt{3}}$$

Rationalize the denominator of \frac{240}{35+35i\sqrt{3}} by multiplying numerator and denominator by 35-35i\sqrt{3}.

$$\frac{240\left(35-35i\sqrt{3}\right)}{\left(35+35i\sqrt{3}\right)\left(35-35i\sqrt{3}\right)}$$

Consider \left(35+35i\sqrt{3}\right)\left(35-35i\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{240\left(35-35i\sqrt{3}\right)}{35^{2}-\left(35i\sqrt{3}\right)^{2}}$$

Calculate 35 to the power of 2 and get 1225.

$$\frac{240\left(35-35i\sqrt{3}\right)}{1225-\left(35i\sqrt{3}\right)^{2}}$$

Expand \left(35i\sqrt{3}\right)^{2}.

$$\frac{240\left(35-35i\sqrt{3}\right)}{1225-\left(35i\right)^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate 35i to the power of 2 and get -1225.

$$\frac{240\left(35-35i\sqrt{3}\right)}{1225-\left(-1225\left(\sqrt{3}\right)^{2}\right)}$$

The square of \sqrt{3} is 3.

$$\frac{240\left(35-35i\sqrt{3}\right)}{1225-\left(-1225\times 3\right)}$$

Multiply -1225 and 3 to get -3675.

$$\frac{240\left(35-35i\sqrt{3}\right)}{1225-\left(-3675\right)}$$

Multiply -1 and -3675 to get 3675.

$$\frac{240\left(35-35i\sqrt{3}\right)}{1225+3675}$$

Add 1225 and 3675 to get 4900.

$$\frac{240\left(35-35i\sqrt{3}\right)}{4900}$$

Divide 240\left(35-35i\sqrt{3}\right) by 4900 to get \frac{12}{245}\left(35-35i\sqrt{3}\right).

$$\frac{12}{245}\left(35-35i\sqrt{3}\right)$$

Use the distributive property to multiply \frac{12}{245} by 35-35i\sqrt{3}.

$$\frac{12}{245}\times 35+\frac{12}{245}\times \left(-35i\right)\sqrt{3}$$

Express \frac{12}{245}\times 35 as a single fraction.

$$\frac{12\times 35}{245}+\frac{12}{245}\times \left(-35i\right)\sqrt{3}$$

Multiply 12 and 35 to get 420.

$$\frac{420}{245}+\frac{12}{245}\times \left(-35i\right)\sqrt{3}$$

Reduce the fraction \frac{420}{245} to lowest terms by extracting and canceling out 35.

$$\frac{12}{7}+\frac{12}{245}\times \left(-35i\right)\sqrt{3}$$

Multiply \frac{12}{245} and -35i to get -\frac{12}{7}i.

$$\frac{12}{7}-\frac{12}{7}i\sqrt{3}$$

Real Part

$$240Re(\frac{1}{35\left(1+\sqrt{3}i\right)})$$