Products and Quotients of Complex Numbers in Polar Form

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Pre-Calculus › Products and Quotients of Complex Numbers in Polar Form

Questions 1 - 10

1

Evaluate:

$(2+i)\cdot(1+4i)$

Explanation

To evaluate this problem we need to FOIL the binomials.

Now recall that

Thus,

2

Evaluate:

$(2+i)\cdot(1+4i)$

Explanation

To evaluate this problem we need to FOIL the binomials.

Now recall that

Thus,

3

Simplify $\frac{3+5i}{2-4i}$ into a number of the form .

$-\frac{7}{10}+\frac{11}{10}i$

$\frac{1}{2}-\frac{3}{2}i$

None of the other answers.

Explanation

We have

$\frac{3+5i}{2-4i}$

Multiply by the complex conjugate of the denominator.

The complex conjugate is the denominator with the sign changed:

$\frac{3+5i}{2-4i}\times \frac{2+4i}{2+4i}$

Multiply fractions

$\frac{(3+5i)(2+4i)}{(2-4i)(2+4i)}$

FOIL the numerator and denominator

$\frac{6+12i+10i+20i^2}{4+8i-8i-16i^2}$

Apply the rule of :

$\frac{6+12i+10i-20}{4+8i-8i+16}$

Simplify.

$\frac{-14+22i}{20}$

Simplify further using the addition of fractions rule, then factor the i out of the 2nd fraction.

$-\frac{7}{10}+\frac{11}{10}i$

4

Simplify: $i^4 \times i^3$

$i+\sqrt{-1}$

Explanation

The expression $i^4 \times i^3$ can be rewritten as:

$i^4 \times i^3 = i^2\times i^2\times i^2\times i$

Since $i=\sqrt{-1}$ , the value of .

$i^2\times i^2\times i^2\times i = (-1) (-1) (-1)(i) = -i$

The correct answer is:

5

Simplify $\frac{3+5i}{2-4i}$ into a number of the form .

$-\frac{7}{10}+\frac{11}{10}i$

$\frac{1}{2}-\frac{3}{2}i$

None of the other answers.

Explanation

We have

$\frac{3+5i}{2-4i}$

Multiply by the complex conjugate of the denominator.

The complex conjugate is the denominator with the sign changed:

$\frac{3+5i}{2-4i}\times \frac{2+4i}{2+4i}$

Multiply fractions

$\frac{(3+5i)(2+4i)}{(2-4i)(2+4i)}$

FOIL the numerator and denominator

$\frac{6+12i+10i+20i^2}{4+8i-8i-16i^2}$

Apply the rule of :

$\frac{6+12i+10i-20}{4+8i-8i+16}$

Simplify.

$\frac{-14+22i}{20}$

Simplify further using the addition of fractions rule, then factor the i out of the 2nd fraction.

$-\frac{7}{10}+\frac{11}{10}i$

6

Simplify: $i^4 \times i^3$

$i+\sqrt{-1}$

Explanation

The expression $i^4 \times i^3$ can be rewritten as:

$i^4 \times i^3 = i^2\times i^2\times i^2\times i$

Since $i=\sqrt{-1}$ , the value of .

$i^2\times i^2\times i^2\times i = (-1) (-1) (-1)(i) = -i$

The correct answer is:

7

Divide.

$\frac{4-i^3}{5+i}$

$\frac{21}{26}+\frac{i}{26}$

$\frac{26}{21}+\frac{26}{i}$

$\frac{21}{26}-\frac{i}{26}$

$\frac{7}{13}-\frac{21i}{6}$

Explanation

To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator.

To find the conjugate, just change the sign in the denominator. The conjugate used will be .

$\left(\frac{4-i^3}{5+i}\right)\left(\frac{5-i}{5-i}\right)$

Now, distribute and simplify.

$\left(\frac{4-i^3}{5+i}\right)\left(\frac{5-i}{5-i}\right)=\frac{20-4i-5i^3+i^4}{25-i^2}$

Recall that $i^2=-1, i^3=-i,\text{ and }i^4=1$

$\frac{20-4i-5i^3+i^4}{25-i^2}=\frac{20-4i-5(-i)+1}{25-(-1)}=\frac{21+i}{26}=\frac{21}{26}+\frac{i}{26}$

8

Divide.

$\frac{4-i^3}{5+i}$

$\frac{21}{26}+\frac{i}{26}$

$\frac{26}{21}+\frac{26}{i}$

$\frac{21}{26}-\frac{i}{26}$

$\frac{7}{13}-\frac{21i}{6}$

Explanation

To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator.

To find the conjugate, just change the sign in the denominator. The conjugate used will be .

$\left(\frac{4-i^3}{5+i}\right)\left(\frac{5-i}{5-i}\right)$

Now, distribute and simplify.

$\left(\frac{4-i^3}{5+i}\right)\left(\frac{5-i}{5-i}\right)=\frac{20-4i-5i^3+i^4}{25-i^2}$

Recall that $i^2=-1, i^3=-i,\text{ and }i^4=1$

$\frac{20-4i-5i^3+i^4}{25-i^2}=\frac{20-4i-5(-i)+1}{25-(-1)}=\frac{21+i}{26}=\frac{21}{26}+\frac{i}{26}$

9

Find the product , if

.

Explanation

To find the product , FOIL the complex numbers. FOIL stands for the multiplication of the Firsts, Outers, Inners, and Lasts.

Using this method we get the following,

and because

.

10

Find the value of ${}(1+i)^{100}$ ,where the complex number is given by ${}i^2=-1$ .

${}-2^{50}$

${}-i2^{100}$

${}i2^{100}$

${}-2^{100}$

${}2^{101}$

Explanation

We note that ${}(1+i)^{2}=1+2i+i^2=1+2i-1=2i$ by FOILing.

We also know that:

$(z)^{nm}=(z^{n})^{m}$

We have by using the above rule: n=2 , m=50

$(1+i)^{100}=((1+i)^{2})^{50}$

Since we know that,

${}(1+i)^{2}=1+2i+i^2=1+2i-1=2i$

We have then:

$(1+i)^{100}=((1+i)^{2})^{50}$ $=(2i)^{50}$

Since we know that:

$(ab)^{n}=a^{n}b^{n}$ , we use a=2 ,b=i

We have then:

$2^{50}({i}^2)^{25}=2^{50}(-1)^{25} =-2^{50}}$

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