Name: Representation of a Complex Number
Uploaded: 2019-07-19
Duration: 6 min 37 s
Description: Learn how to express a complex number and represent it geometrically. Also, understand the concept of modulus and the argument of a complex number.

Question

Express the following in the form $a + i b, a, b \in R$ , using De Moivre's theorem.

$(1 - i)^{5}$

Accepted Answer

Let $z = 1 - i = r (\cos θ + i \sin θ)$

This is of the $a + b i$ , where $a = 1, b = - 1$

$∴ r = \sqrt{a^{2} + b^{2}}$

$= \sqrt{1^{2} + (- 1)^{2}} = \sqrt{2}$

Also, $\cos θ = \frac{a}{r} = \frac{1}{\sqrt{2}}$ and $\sin θ = \frac{b}{r} = \frac{- 1}{\sqrt{2}}$

$∴ θ$ lies in the fourth quadrant.

$∴ \tan θ = \frac{\sin θ}{\cos θ} = \frac{(\frac{- 1}{\sqrt{2}})}{(\frac{1}{\sqrt{2}})} = - 1$

$= - \tan \frac{π}{4} = \tan (2 π - \frac{π}{4})$

$∴ θ = \frac{7 π}{4}$

$∴ z = \sqrt{2} (\cos \frac{7 π}{4} + i \sin \frac{7 π}{4})$

$∴ z^{5} = (1 - i)^{5}$

$= {[\sqrt{2} (\cos \frac{7 π}{4} + i \sin \frac{7 π}{4})]}^{5}$

$= (\sqrt{2})^{5} {[\cos \frac{7 π}{4} + i \sin \frac{7 π}{4}]}^{5}$

$= 4 \sqrt{2} (\cos \frac{35 π}{4} + i \sin \frac{35 π}{4})$ $[∵ (\cos θ + i \sin θ)^{n} = \cos n θ + i \sin n θ]$

$= 4 \sqrt{2} [\cos (9 π - \frac{π}{4}) + i \sin (9 π - \frac{π}{4})]$

$= 4 \sqrt{2} (- \cos \frac{π}{4} + i \sin \frac{π}{4})$

$= 4 \sqrt{2} (- \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}})$

$= - 4 + 4 i$

Hence, given expression can be written in the form $a + i b$ . Here $a = - 4 and b = 4$ .

Express the following in the form $a + i b, a, b \in R$ , using De Moivre's theorem.

$(1 - i)^{5}$

Important Questions on Complex Numbers

Learn and Prepare for any exam you want !

Express the following in the form a+ib, a, b∈R, using De Moivre's theorem. (1-i)5

Important Questions on Complex Numbers

Learn and Prepare for any exam you want !

Express the following in the form $a + i b, a, b \in R$ , using De Moivre's theorem.

$(1 - i)^{5}$