Solve frac{-3sqrt{5}+2i}{2+3sqrt{5}i}

$$\frac { - 3 \sqrt { 5 } + 2 i } { 2 + 3 \sqrt { 5 } i }$$

Evaluate

$$i$$

Solution Steps

Rationalize the denominator of \frac{-3\sqrt{5}+2i}{2+3i\sqrt{5}} by multiplying numerator and denominator by 2-3i\sqrt{5}.

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

Consider \left(2+3i\sqrt{5}\right)\left(2-3i\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(-3\sqrt{5}+2i\right)\left(2-3i\sqrt{5}\right)}{2^{2}-\left(3i\sqrt{5}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(-3\sqrt{5}+2i\right)\left(2-3i\sqrt{5}\right)}{4-\left(3i\sqrt{5}\right)^{2}}$$

Expand \left(3i\sqrt{5}\right)^{2}.

$$\frac{\left(-3\sqrt{5}+2i\right)\left(2-3i\sqrt{5}\right)}{4-\left(3i\right)^{2}\left(\sqrt{5}\right)^{2}}$$

Calculate 3i to the power of 2 and get -9.

$$\frac{\left(-3\sqrt{5}+2i\right)\left(2-3i\sqrt{5}\right)}{4-\left(-9\left(\sqrt{5}\right)^{2}\right)}$$

The square of \sqrt{5} is 5.

$$\frac{\left(-3\sqrt{5}+2i\right)\left(2-3i\sqrt{5}\right)}{4-\left(-9\times 5\right)}$$

Multiply -9 and 5 to get -45.

$$\frac{\left(-3\sqrt{5}+2i\right)\left(2-3i\sqrt{5}\right)}{4-\left(-45\right)}$$

Multiply -1 and -45 to get 45.

$$\frac{\left(-3\sqrt{5}+2i\right)\left(2-3i\sqrt{5}\right)}{4+45}$$

Add 4 and 45 to get 49.

$$\frac{\left(-3\sqrt{5}+2i\right)\left(2-3i\sqrt{5}\right)}{49}$$

Apply the distributive property by multiplying each term of -3\sqrt{5}+2i by each term of 2-3i\sqrt{5}.

$$\frac{-6\sqrt{5}+9i\left(\sqrt{5}\right)^{2}+4i+6\sqrt{5}}{49}$$

The square of \sqrt{5} is 5.

$$\frac{-6\sqrt{5}+9i\times 5+4i+6\sqrt{5}}{49}$$

Multiply 9i and 5 to get 45i.

$$\frac{-6\sqrt{5}+45i+4i+6\sqrt{5}}{49}$$

Add 45i and 4i to get 49i.

$$\frac{-6\sqrt{5}+49i+6\sqrt{5}}{49}$$

Combine -6\sqrt{5} and 6\sqrt{5} to get 0.

$$\frac{49i}{49}$$

Divide 49i by 49 to get i.

$$i$$

Real Part

$$0$$

Solution Steps

Rationalize the denominator of \frac{-3\sqrt{5}+2i}{2+3i\sqrt{5}} by multiplying numerator and denominator by 2-3i\sqrt{5}.

$$Re(\frac{\left(-3\sqrt{5}+2i\right)\left(2-3i\sqrt{5}\right)}{\left(2+3i\sqrt{5}\right)\left(2-3i\sqrt{5}\right)})$$

Consider \left(2+3i\sqrt{5}\right)\left(2-3i\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$Re(\frac{\left(-3\sqrt{5}+2i\right)\left(2-3i\sqrt{5}\right)}{2^{2}-\left(3i\sqrt{5}\right)^{2}})$$

Calculate 2 to the power of 2 and get 4.

$$Re(\frac{\left(-3\sqrt{5}+2i\right)\left(2-3i\sqrt{5}\right)}{4-\left(3i\sqrt{5}\right)^{2}})$$

Expand \left(3i\sqrt{5}\right)^{2}.

$$Re(\frac{\left(-3\sqrt{5}+2i\right)\left(2-3i\sqrt{5}\right)}{4-\left(3i\right)^{2}\left(\sqrt{5}\right)^{2}})$$

Calculate 3i to the power of 2 and get -9.

$$Re(\frac{\left(-3\sqrt{5}+2i\right)\left(2-3i\sqrt{5}\right)}{4-\left(-9\left(\sqrt{5}\right)^{2}\right)})$$

The square of \sqrt{5} is 5.

$$Re(\frac{\left(-3\sqrt{5}+2i\right)\left(2-3i\sqrt{5}\right)}{4-\left(-9\times 5\right)})$$

Multiply -9 and 5 to get -45.

$$Re(\frac{\left(-3\sqrt{5}+2i\right)\left(2-3i\sqrt{5}\right)}{4-\left(-45\right)})$$

Multiply -1 and -45 to get 45.

$$Re(\frac{\left(-3\sqrt{5}+2i\right)\left(2-3i\sqrt{5}\right)}{4+45})$$

Add 4 and 45 to get 49.

$$Re(\frac{\left(-3\sqrt{5}+2i\right)\left(2-3i\sqrt{5}\right)}{49})$$

Apply the distributive property by multiplying each term of -3\sqrt{5}+2i by each term of 2-3i\sqrt{5}.

$$Re(\frac{-6\sqrt{5}+9i\left(\sqrt{5}\right)^{2}+4i+6\sqrt{5}}{49})$$

The square of \sqrt{5} is 5.

$$Re(\frac{-6\sqrt{5}+9i\times 5+4i+6\sqrt{5}}{49})$$

Multiply 9i and 5 to get 45i.

$$Re(\frac{-6\sqrt{5}+45i+4i+6\sqrt{5}}{49})$$

Add 45i and 4i to get 49i.

$$Re(\frac{-6\sqrt{5}+49i+6\sqrt{5}}{49})$$

Combine -6\sqrt{5} and 6\sqrt{5} to get 0.

$$Re(\frac{49i}{49})$$

Divide 49i by 49 to get i.

$$Re(i)$$

The real part of i is 0.

$$0$$