Answered: "How do you simplify |2+3i|?"

Secondary
Algebra
Algebra II
Complex numbers

Answered question

2022-01-17

How do you simplify

| 2 + 3 i | ?

Answer & Explanation

nick1337

Expert2022-01-18Added 573 answers

Step 1 Given:

| 2 + 3 i |

| a + b i | = \sqrt{a^{2} + b^{2}}

So,

| 2 + 3 i | = \sqrt{2^{2} + 3^{2}}

\Rightarrow \sqrt{4 + 9}

\Rightarrow \sqrt{13}

Hence,

| 2 + 3 i | = \sqrt{13}

Vasquez

Skilled2022-01-18Added 457 answers

Step 1 The inverse of

2 + 3 i

\frac{1}{2 + 3 i}

In general case, multiply the expression

\frac{1}{a + b i}

by the conjugate (the conjugate of

a + i b

a - i b

\frac{1}{a + i b} = \frac{1}{(a - i b) (a + i b)} (a - i b)

Expand the denominator:

\frac{1}{(a - i b) (a + i b)} (a - i b) = \frac{a - i b}{a^{2} + b^{2}}

Split:

\frac{a - i b}{a^{2} + b^{2}} = \frac{a}{a^{2} + b^{2}} - \frac{i b}{a^{2} + b^{2}}

In our case,

a = 2

and

b = 3

Therefore,

(\frac{1}{2 + 3 i}) = (\frac{2}{13} - \frac{3 i}{13})

Hence,

\frac{1}{2 + 3 i} = \frac{2}{13} - \frac{3 i}{13}

alenahelenash

Skilled2022-01-24Added 366 answers

Step 1 We have that

a = 2

and

b = 3

Thus,

r = \sqrt{(2)^{2} + (3)^{2}} = \sqrt{13}

Also,

θ = a \tan (\frac{3}{2}) = a \tan (\frac{3}{2})

Therefore,

2 + 3 i = \sqrt{13} (\cos (a \tan (\frac{3}{2})) + i \sin (a \tan (\frac{3}{2})))

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