Question

reduce (1/1+2i+3/1-i)*(3-2i/1+3i) to the form a+ib​

Answer 1

The a + ib format is 0.7 + i(2.2).

• Complex numbers are those that are expressed as a+ib, where a and b are actual numbers and I is a fictitious number called a "iota." I has the value root of  (-1). Typically, the letters I or "j," which are equivalent to root of -1, are used to denote imaginary numbers. As a result, the imaginary number's square has a negative value. A real number and an imaginary number are effectively combined to create a complex number. The complex number is written as a+ib, where a and ib are real and imaginary numbers, respectively.
• 0 is a real number, as is well known. Additionally, complex numbers include real numbers. Consequently, 0 is a complex number as well and can be written as 0+0i.

Here, according to the given information, we are given that,

$(\frac{1}{1+2i} +\frac{3}{1-i} ).(\frac{3-2i}{1+3i} )\\=(\frac{1-2i}{5} +\frac{3(1+i)}{2} ).(\frac{(3-2i)(1-3i)}{10} )\\\\=(\frac{2-4i+15i+15}{10} ).(\frac{3-9i-2i-6}{10} )\\=(\frac{17+11i}{10}) .(\frac{-3-11i}{10} )\\=\frac{-51-187i-33i+121}{100} \\=\frac{70-220i}{100} \\=0.7+(2.2i)$

Hence, the a + ib format is 0.7 + i(2.2).

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