Answer:
To divide the complex number, we multiply top and bottom of the fraction by the conjugate of the denominator.
Given the complex number: [tex]\frac{2+2i}{1-i}[/tex]
Multiply the conjugate of the denominator (1 + i) to the top and bottom.
we have;
[tex]\frac{(2+2i) \cdot (1+i)}{(1-i)\cdot (1+i)}[/tex]
Using distributive property; [tex]a\cdot (b+c) = a\cdot b+ a\cdot c[/tex]
[tex]\frac{2+2i+2i+2i^2}{1-i^2} = \frac{2+4i-2}{1-(-1)} = \frac{0+4i}{1+1} = \frac{4i}{2}= 2i[/tex] [ Since, [tex]i^2 = -1[/tex] ]
Therefore, the standard form of this complex number [tex]\frac{2+2i}{1-i}[/tex] is; 0 + 2i
Answer:
[tex]0+2i[/tex]
Step-by-step explanation:
The given complex number is
[tex]\frac{2+2i}{1-i} =\frac{2(1+i)}{1-i}[/tex]
We multiply by the conjugate of the denominator to get,
[tex]=\frac{2(1+i)(1+i)}{(1-i)(1+i)}[/tex]
We apply difference of two squares and perfect squares to obtain,
[tex]=\frac{2(1+2i+i^2)}{(1^2-i^2)}[/tex]
[tex]=\frac{2(1+2i+-1)}{(1--1)}[/tex]
[tex]=\frac{2(1+2i-1)}{(1+1)}[/tex]
[tex]=\frac{2(2i)}{(2)}=2i[/tex]
In standard form we rewrite as [tex]a+bi[/tex]
The given complex number in standard form is
[tex]0+2i[/tex]
