Solve i+1/5:sqrt{(sqrt{15})^2+(-sqrt{10})^2}-|-4|=

$$i + \frac { 1 } { 5 } : \sqrt { ( \sqrt { 15 } ) ^ { 2 } + ( - \sqrt { 10 } ) ^ { 2 } } - | - 4 | =$$

Evaluate

$$-\frac{99}{25}+i=-3.96+i$$

Solution Steps

The square of \sqrt{15} is 15.

$$i+\frac{\frac{1}{5}}{\sqrt{15+\left(-\sqrt{10}\right)^{2}}}-|-4|$$

Calculate -\sqrt{10} to the power of 2 and get \left(\sqrt{10}\right)^{2}.

$$i+\frac{\frac{1}{5}}{\sqrt{15+\left(\sqrt{10}\right)^{2}}}-|-4|$$

The square of \sqrt{10} is 10.

$$i+\frac{\frac{1}{5}}{\sqrt{15+10}}-|-4|$$

Add 15 and 10 to get 25.

$$i+\frac{\frac{1}{5}}{\sqrt{25}}-|-4|$$

Calculate the square root of 25 and get 5.

$$i+\frac{\frac{1}{5}}{5}-|-4|$$

Express \frac{\frac{1}{5}}{5} as a single fraction.

$$i+\frac{1}{5\times 5}-|-4|$$

Multiply 5 and 5 to get 25.

$$i+\frac{1}{25}-|-4|$$

The modulus of a complex number a+bi is \sqrt{a^{2}+b^{2}}. The modulus of -4 is 4.

$$i+\frac{1}{25}-4$$

Do the additions.

$$-\frac{99}{25}+i$$

Real Part

$$-\frac{99}{25} = -3\frac{24}{25} = -3.96$$

Solution Steps

The square of \sqrt{15} is 15.

$$Re(i+\frac{\frac{1}{5}}{\sqrt{15+\left(-\sqrt{10}\right)^{2}}}-|-4|)$$

Calculate -\sqrt{10} to the power of 2 and get \left(\sqrt{10}\right)^{2}.

$$Re(i+\frac{\frac{1}{5}}{\sqrt{15+\left(\sqrt{10}\right)^{2}}}-|-4|)$$

The square of \sqrt{10} is 10.

$$Re(i+\frac{\frac{1}{5}}{\sqrt{15+10}}-|-4|)$$

Add 15 and 10 to get 25.

$$Re(i+\frac{\frac{1}{5}}{\sqrt{25}}-|-4|)$$

Calculate the square root of 25 and get 5.

$$Re(i+\frac{\frac{1}{5}}{5}-|-4|)$$

Express \frac{\frac{1}{5}}{5} as a single fraction.

$$Re(i+\frac{1}{5\times 5}-|-4|)$$

Multiply 5 and 5 to get 25.

$$Re(i+\frac{1}{25}-|-4|)$$

The modulus of a complex number a+bi is \sqrt{a^{2}+b^{2}}. The modulus of -4 is 4.

$$Re(i+\frac{1}{25}-4)$$

Do the additions in i+\frac{1}{25}-4.

$$Re(-\frac{99}{25}+i)$$

The real part of -\frac{99}{25}+i is -\frac{99}{25}.

$$-\frac{99}{25}$$