Question

complex no .write in standard form a+ib (1/3+3i)^3

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{(a+\iota b)=\frac{-53}{27}-26\iota}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Complex \: number= (\frac{1}{3} + 3 \iota)^{2} \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Standard \: form(a + \iota b) = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies a + \iota b = {(\frac{1}{3} + 3 \iota)}^{3} \\\\ \tt\circ\:(a+b)^{3}=a^{3}+b^{3}+3a^{2}b+3ab^{2}\\ \\ \tt: \implies a + \iota b = (\frac{1}{3})^{3} + {(3 \iota)}^{3} + 3 \times (\frac{1}{3})^{2} \times 3 \iota + 3 \times \frac{1}{3} \times { (3 \iota)}^{2} \\ \\ \tt: \implies a + \iota b = \frac{1}{27} + 27 \times { \iota}^{3} + 3 \times \frac{1}{9} \times 3 \iota + 3 \times { \iota}^{2} \\ \\ \tt: \implies a + \iota b = \frac{1}{27} + 27 \times - \iota + \iota + 3 \times - 1 \\ \\ \tt: \implies a + \iota b = \frac{1}{27} - 27 \iota + \iota - 3 \\ \\ \tt: \implies a + \iota b = \frac{1 - 54}{27} - 26 \iota \\ \\ \green{\tt: \implies a + \iota b = \frac{ - 53}{27} - 26 \iota}$

Answer 2

Given ,

The complex number is (1/3 + 3i)³

We know that ,

The standard equation of complex number is a + ib

Thus ,

$\tt \implies {( \frac{1}{3} + 3i )}^{3}$

$\tt \implies {( \frac{1}{3}) }^{2} + {(3i)}^{3} + 3. {( \frac{1}{3} )}^{2} .3i + 3. \frac{1}{2}. {(3i)}^{2}$

$\tt \implies \frac{1}{27} + 27 {(i)}^{3} + \frac{9}{9}i + 9 {(i)}^{2}$

$\tt \implies \frac{1}{27} + 27( - i) + i + 9( - 1)$

$\tt \implies \frac{1}{27} - 27i + i - 9$

$\tt \implies ( \frac{1}{27} - 9) + i( - 27 + 1)$

$\tt \implies - \frac{242}{27} - 26i$

Remmember :

$\tt \mapsto (a + b)³ = (a)³ + (b)³ + 3(a)²b + 3a(b)²$

$\tt \mapsto (i)² = -1$

$\tt \mapsto (i)³ = -(i)$