What is ((1+2i)/(1-3i))^2 ?

The solution to ((1+2i)/(1-3i))^2 is -1/2 i

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simplify ((1+2i)/(1-3i))^2

simplify

(\frac{1 + 2 i}{1 - 3 i})^{2}

- \frac{1}{2} i

Solution steps

(\frac{1 + 2 i}{1 - 3 i})^{2}

Simplify

\frac{1 + 2 i}{1 - 3 i} : - \frac{1}{2} + \frac{1}{2} i

= (- \frac{1}{2} + \frac{1}{2} i)^{2}

\frac{1}{2} i = \frac{i}{2}

= (- \frac{1}{2} + \frac{i}{2})^{2}

Combine the fractions

- \frac{1}{2} + \frac{i}{2} : \frac{- 1 + i}{2}

= (\frac{- 1 + i}{2})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( - 1 + i ) ^{2}}{2 ^{2}}

2^{2} = 4

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

Expand

(- 1 + i)^{2} : - 2 i

= \frac{- 2 i}{4}

Factor the number:

4 = 2 \cdot 2

= \frac{- 2 i}{2 \cdot 2}

Cancel the common factor:

2

= \frac{- i}{2}

Rewrite in standard complex form:

- \frac{1}{2} i

= - \frac{1}{2} i

- 86 + 20 x \leq 4 + 2 x

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- \frac{6}{3}

\frac{( - 13 125 )}{( 5 5 )}

7 x^{2} + 5 x - 2 = 0