Simplify and write in standard form (1 + 5i)/-3i

Last Updated : 26 Dec, 2023

Complex numbers are those with the formula a + ib, where a and b are real numbers and I (iota) is the imaginary component and represents (-1), and are often represented in rectangle or standard form. 10 + 5i, for example, is a complex number in which 10 represents the real component and 5i represents the imaginary part. Depending on the values of a and b, they might be wholly real or purely fictitious. When a = 0 in a + ib, ib is a totally imaginary number, and when b = 0, we get a, which is a strictly real number.

Dividing Two Complex Numbers

The division process of two complex numbers is slightly different from that of the division process of two real numbers. Dividing complex numbers is more like the concept of rationalizing the denominator in the case of fractions involving irrational numbers as their denominators.

The following steps are involved:

Assure that both the numerator and numerator are in the standard form of complex numbers, i.e., z = a + ib.
Calculate the complex conjugate of the denominator. Say, if the denominator is c + id, then its conjugate is c − id.
Multiply the conjugate with both the terms of the fraction.
Use the difference of squares formula to solve the denominator.
Divide the obtained complex number into its real and imaginary parts.

The division process of two complex numbers z₁ = x + iy and z₂ = a + ib is shown as follows:

$\begin{aligned}\dfrac{z_1}{z_2}&=\dfrac{x+iy}{a+ib}\\&=\dfrac{x+iy}{a+ib}\times\dfrac{a-ib}{a-ib}\\&=\dfrac{(x+iy)(a-ib)}{a^2-(ib)^2}\\&=\dfrac{ax-ibx+iay-i^2by}{a^2-(-1)b^2}\\&=\dfrac{ac-ibx+iay+by}{a^2+b^2}\\&=\dfrac{(ax+by)+i(ay-bx)}{a^2+b^2}\\&=\dfrac{ax+by}{a^2+b^2}+i\left(\dfrac{ay-bx}{a^2+b^2}\right)\end{aligned}$

Simplify and write the answer in standard form (1 + 5i)/-3i.

Solution:

Standard form of denominator −3i = 0 − 3i

Conjugate of the denominator = 0 + 3i

Multiply both the numerator and denominator by 0 + 3i.

$\begin{aligned}\dfrac{z_1}{z_2}&=\dfrac{1+5i}{0-3i}\\&=\dfrac{1+5i}{0-3i}\times\dfrac{0+3i}{0+3i}\\&=\dfrac{(1+5i)(0+3i)}{0^2-(3i)^2}\\&=\dfrac{3i+15i^2}{0^2-(-1)3^2}\\&=\dfrac{3i-15}{0^2+3^2}\\&=\dfrac{3i-15}{9}\\&=\dfrac{-5}{3}+i\left(\dfrac{1}{3}\right)\end{aligned}$

Similar Problems

Problem 1. Solve: $\frac{-3-5i}{2i}$ .

Solution:

Standard form of denominator 2i = 0 + 2i

Conjugate of the denominator = 0 − 2i

Multiply both the numerator and denominator by 0 + 2i.

$\begin{aligned}\dfrac{z_1}{z_2}&=\dfrac{-3-5i}{0+2i}\\&=\dfrac{-3-5i}{0+2i}\times\dfrac{0-2i}{0-2i}\\&=\dfrac{(-3-5i)(0-2i)}{0^2-(2i)^2}\\&=\dfrac{-5}{2}+i\left(\dfrac{3}{2}\right)\end{aligned}$

Problem 2. Solve: $\frac{-1}{3+2i}$ .

Solution:

Conjugate of the denominator = 3 + 2i

Multiply both the numerator and denominator by 3 – 2i.

$\begin{aligned}\dfrac{z_1}{z_2}&=\dfrac{-1}{3+2i}\\&=\dfrac{-1}{3-2i}\times\dfrac{3+2i}{3+2i}\\&=\dfrac{(-3+2i)}{3^2-(2i)^2}\\&=\dfrac{-3}{13}+i\left(\dfrac{2}{13}\right)\end{aligned}$

Problem 3. Solve: $\frac{1}{3+2i}$ .

Solution:

Conjugate of the denominator = 3 + 2i

Multiply both the numerator and denominator by 3 – 2i.

$\begin{aligned}\dfrac{z_1}{z_2}&=\dfrac{1}{3+2i}\\&=\dfrac{1}{3-2i}\times\dfrac{3+2i}{3+2i}\\&=\dfrac{(3+2i)}{3^2-(2i)^2}\\&=\dfrac{3}{13}+i\left(\dfrac{2}{13}\right)\end{aligned}$

Problem 4. Solve (5+√2i)/(1−√2i).

Solution:

Conjugate of denominator = 1 + √2i

Multiply both the numerator and denominator by 1 + √2i.

$\begin{aligned}\dfrac{5+\sqrt{2}i}{1-\sqrt{2}i}=\dfrac{a+ib}{c+id}&=\dfrac{ac+bd}{c^2+d^2}+i\left(\dfrac{bc-ad}{c^2+d^2}\right)\\&=\dfrac{(5\times 1)+(\sqrt{2}\times(-\sqrt{2}))}{1^2+(-\sqrt{2})^2}+\\&i\left(\dfrac{(\sqrt{2} \times 1)-(5\times(-\sqrt{2}))}{1^2+(-\sqrt{2})^2}\right)\\&=\dfrac{5-2}{1+2}+i\left(\dfrac{\sqrt{2}+5\sqrt{2}}{1+2}\right)\\&=\dfrac{3+6\sqrt{2}i}{1+2}\\&=1+2\sqrt{2}i\end{aligned}$

Problem 5. Solve: $\frac{1-5i}{-3i}$ .

Solution:

Standard form of denominator −3i = 0 − 3i

Conjugate of the denominator = 0 + 3i

Multiply both the numerator and denominator by 0 + 3i.

$\begin{aligned}\dfrac{z_1}{z_2}&=\dfrac{1-5i}{0-3i}\\&=\dfrac{1-5i}{0-3i}\times\dfrac{0+3i}{0+3i}\\&=\dfrac{(1-5i)(0+3i)}{0^2-(3i)^2}\\&=\dfrac{3i-15i^2}{0^2-(-1)3^2}\\&=\dfrac{3i+15}{0^2+3^2}\\&=\dfrac{3i+15}{9}\\&=\dfrac{5}{3}+i\left(\dfrac{1}{3}\right)\end{aligned}$

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Simplify and write in standard form (1 + 5i)/-3i

Dividing Two Complex Numbers

Simplify and write the answer in standard form (1 + 5i)/-3i.

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