Reduce 1-1-2 i -3-1- i 3-2 i -1-3 i to the form a-ib-

We have ( 1/1+2i+3/1 i ) ( 3 2i/1+3i ) = {(1 i)+(3+6i)/(1+2i)(1 i)} ( 3 2i/1+3i ) = (4+5i)/(3+i)×(3 2i)/(1+3i) =(4+5i)(3 2i)/(3+i)(1+3i) =(12+10)+(15 8)i/(3 3 ...

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Byju's Answer

Standard XI

Mathematics

Sum of Infinite Terms of a GP

Reduce 1/1+2 ...

Question

Reduce $(\frac{1}{1 + 2 i} + \frac{3}{1 - i}) (\frac{3 - 2 i}{1 + 3 i})$ to the form (a+ib).

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Solution

We have

$(\frac{1}{1 + 2 i} + \frac{3}{1 - i}) (\frac{3 - 2 i}{1 + 3 i}) = {\frac{(1 - i) + (3 + 6 i)}{(1 + 2 i) (1 - i)}} (\frac{3 - 2 i}{1 + 3 i}) = \frac{(4 + 5 i)}{(3 + i)} \times \frac{(3 - 2 i)}{(1 + 3 i)}$

$= \frac{(4 + 5 i) (3 - 2 i)}{(3 + i) (1 + 3 i)} = \frac{(12 + 10) + (15 - 8) i}{(3 - 3) + (9 i + i)} = \frac{(22 + 7 i)}{10 i} \times \frac{i}{i}$

$= \frac{(7 i^{2} + 22 i)}{10 i^{2}} = \frac{- 7 + 22 i}{- 10} = (\frac{7}{10} - \frac{22}{10} i) = (\frac{7}{10} - \frac{11}{5} i)$ .

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Express the following complex numbers in the standard form a + i b :

$(i) (1 + i) (1 + 2 i) (i i) \frac{3 + 2 i}{- 2 + i} (i i i) \frac{1}{(2 + i)^{2}} (i v) \frac{1 - i}{1 + i} (v) \frac{(2 + i)^{3}}{2 + 3 i} (v i) \frac{(1 + i) (1 + \sqrt{3 i})}{1 - i} (v i i) \frac{2 + 3 i}{4 + 5 i} (v i i i) \frac{(1 - i)^{3}}{1 - i^{3}} (i x) (1 + 2 i)^{- 3} (x) \frac{3 - 4 i}{(4 - 2 i) (1 + i)} (x i) (\frac{1}{1 - 4 i} - \frac{2}{1 + i}) (\frac{3 - 4 i}{5 + i}) (x i i) \frac{5 + \sqrt{2 i}}{1 - \sqrt{2 i}}$