Question

PLEASE HELP Multiply the following complex numbers. Reduce terms and simplify. Explain how your simplified result and the first term in the pair below are related algebraically to each other and to the complex number (1 + i).
(IN THE PICTURE BELOW IT HAS A TIMES SYMBOLE BETWEEN)

Answer 1

SOS

Answer:

$=\frac{1}{2}-\frac{1}{2}i$

Step-by-step explanation:

*i is the imaginary number

$\frac{1-i}{1-i}=1$

$\mathrm{Multiply:}\:\frac{1}{1+i}\cdot \:1=\frac{1}{1+i}$

$\mathrm{Rationalize\:}\frac{1}{1+i}:\quad \frac{1-i}{2}$

$\mathrm{Rewrite\:}\frac{1-i}{2}\mathrm{\:in\:standard\:complex\:form:\:}\frac{1}{2}-\frac{1}{2}i$

$=\frac{1}{2}-\frac{1}{2}i$

Hope this helps!!!

Answer 2

After the simplification of $\frac{1}{1+i}$ we get  $\frac{1}{2} -\frac{1}{2}i$.

What is the rationalization of complex number?

The denominator can be forced to be real by multiplying both numerator and denominator by the conjugate of the denominator is called rationalization of complex number.

According to the given question.

We have a complex number.

$\frac{1}{1+i}$

For the simplification of the above complex number we will do rationalization.

$\frac{1}{1+i}$

$=\frac{1}{1+i} (\frac{1-i}{1-i})$                           (conjugate of $1+ i$ is $1-i$)

$=\frac{1-i}{(1)^{2}-(i)^{2} }$                       (by using the property $(a-b)(a+b) =(a^{2} -b^{2})$)

$=\frac{1-i}{1-(-1)}$                           ( because $(i)^{2} = -1$ )

$=\frac{1-i}{1+1}$

$=\frac{1-i}{2}$

$=\frac{1}{2} -\frac{1}{2}i$

Hence, $\frac{1}{1+i}$ is equals to $\frac{1}{2} -\frac{1}{2}i$.

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