Question

Sums and Differences Evaluate the sum or difference, and write the result in the form $a+bi$.

$6i-\left(4-i\right)$

Short answer

The difference of complex numbers is $-4+7i$.

Step-by-step solution

Step 1. Concept of complex number.

A complex number is of the form $a+bi$. Where a and b are some real numbers and i is the imaginary number.

The number a is known as the real part of the complex number and b is known as imaginary part of the complex number.

Step 2. Concept of adding/subtracting complex numbers.

To add or subtract two complex numbers, simply add/subtract the real parts together and the imaginary parts together. That is,

$\left(a+bi\right)+\left(c+di\right)=\left(a+c\right)+\left(b+d\right)i$

In the complex number $2+5i$, the real part is 2 and the complex part is 5.

The imaginary number i is defined as the square root of negative one. That is, $i=\sqrt{-1}$.

Step 3. Find the sum/difference.

To find the difference of the given complex numbers, first rewrite both complex numbers in standard form and then subtract the real parts together and complex parts together as shown below

$$\begin{gathered} 6i-\left(4-i\right)=\left(0+6i\right)-\left(4-i\right) \\ =\left(0-4\right)+\left(6-\left(-1\right)\right)i \\ =\left(-4\right)+\left(6+1\right)i \\ =-4+7i \end{gathered}$$

Thus, the difference of the complex numbers is $-4+7i$.