polar(2+2i) — step-by-step solution

Let's write

2 + 2 i

in polar form:

r \exp (i θ)

, where r is the modulus and θ is an argument

The modulus of the complex number

a + i b

\sqrt{a^{2} + b^{2}}

thus:

ⅰ

\mod (2 + 2 i)

\sqrt{2^{2} + 2^{2}}

\sqrt{2^{2} + 2^{2}}

ⓘ Calculate powers before multiplications, additions and subtractions. With a positive integer exponent, a power corresponds to a repeated product:

2^{2}

2 \times 2

4

■Powers are computed before multiplications, additions and subtractions. A power with a positive integer exponent is a repeated product:

2^{2}

2 \times 2

4

\sqrt{4 + 4}

ⓘ Start by calculating what is inside the parentheses:

4 + 4

8

\sqrt{8}

ⓘ Let's look for the largest square dividing

8

4

2^{2}

, and take it out of the root:

\sqrt{8}

\sqrt{4 \times 2}

\sqrt{4} \times \sqrt{2}

2 \times \sqrt{2}

2 \times \sqrt{2}

For a nonzero complex number

a + i b

, the argument is the directed angle from the positive real axis to the vector

(a, b)

; depending on the signs of a and b, it is

0

π

\frac{π}{2}

- \frac{π}{2}

atan (\frac{b}{a})

atan (\frac{b}{a}) + π

atan (\frac{b}{a}) - π

ⅰ

\arg (2 + 2 i)

\arctan (\frac{2}{2})

\arctan (\frac{2}{2})

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{2}{2}

1

\arctan (1)

ⓘ acos, asin and atan give the principal angle whose cosine, sine or tangent respectively equals the number, here a remarkable value:

\arctan (1)

\frac{π}{4}

\frac{π}{4}

So the polar form is:

ⅰ

polar (2 + 2 i)

2 \times \sqrt{2} \times \exp (i \times \frac{π}{4})

2 \times \sqrt{2} \times \exp (i \times \frac{π}{4})

ⓘ Multiplying a fraction by a term multiplies its numerator:

i \times \frac{π}{4}

\frac{i π}{4}

2 \times \sqrt{2} \times \exp (\frac{i π}{4})

▶Result:

2 \exp (\frac{i π}{4}) \times \sqrt{2}