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Home » Express the following complex numbers in the standard form a + ib: (i) (ii)
Express the following complex numbers in the standard form a + ib:
(i)\left(\frac{1}{1-4 i}-\frac{2}{1+i}\right)\left(\frac{3-4 i}{5+i}\right)
(ii) (5+\sqrt{2 i}) /(1-\sqrt{2 i})
CBSE | Class 11 | Complex Numbers | Exercise 13.2 | Maths | RD Sharma
Express the following complex numbers in the standard form a + ib:
(i)\left(\frac{1}{1-4 i}-\frac{2}{1+i}\right)\left(\frac{3-4 i}{5+i}\right)
(ii) (5+\sqrt{2 i}) /(1-\sqrt{2 i})

Solution:

(i)\left(\frac{1}{1-4 i}-\frac{2}{1+i}\right)\left(\frac{3-4 i}{5+i}\right)
Simplify and express in the standard form of (a + ib),
\begin{aligned} \left(\frac{1}{1-4 i}-\frac{2}{1+i}\right)\left(\frac{3-4 i}{5+i}\right) =\left(\frac{1+i-2(1-4 i)}{(1-4 i)(1+i)}\right)\left(\frac{3-4 i}{5+i}\right) \\ =\left(\frac{1+i-2+8 i}{1(1+i)-4 i(1+i)}\right)\left(\frac{3-4 i}{5+i}\right) \end{aligned}
\begin{aligned} =\left(\frac{-1+9 i}{1+i-4 i-4 i^{2}}\right)\left(\frac{3-4 i}{5+i}\right) \\ =\left(\frac{-1+9 i}{1-3 i-4(-1)}\right)\left(\frac{3-4 i}{5+i}\right) \\ =\frac{(-1+9 i)(3-4 i)}{(5-3 i)(5+i)} \\ =\frac{-1(3-4 i)+9 i(3-i)}{5(5+i)-3 i(5+i)} \\ =\frac{-3+4 i+27 i-9 i^{2}}{25+5 i-15 i-3 i^{2}} \\ =\frac{-3+31 i-9(-1)}{25-10 i-3(-1)} \\ =\frac{6+31 i}{28-10 i} \end{aligned}
Multiplying and dividing with (28+10 \mathrm{i})
\begin{aligned} =\frac{6+31 i}{28-10 i} \times \frac{28+10 i}{28+10 i} \\ =\frac{6(28+10 i)+31 i(28+10 i)}{28^{2}-(10 i)^{2}} \\ =\frac{168+60 i+868 i+310 i^{2}}{794-100 i^{2}} \\ =\frac{168+928 i+310(-1)}{784-100(-1)} \\ =\frac{478+928 i}{884} \end{aligned}
As a result, the values of a, b are 478 / 884,928 / 884

(ii) (5+\sqrt{2 i}) /(1-\sqrt{2 i})
Simplify and express in the standard form of (a + ib),
(5+\sqrt{2 i}) /(1-\sqrt{2 i})
Multiplying and dividing with (1+\sqrt{2 i})]
=(5+\sqrt{2 i}) /(1-\sqrt{2 i}) \times(1+\sqrt{2 i}) /(1+\sqrt{2 i})

=[5(1+\sqrt{2} i)+\sqrt{2 i}(1+\sqrt{2 i})] /\left(1^{2}-(\sqrt{2})^{2}\right)
=\left[5+5 \sqrt{2 i}+\sqrt{2 i}+2 i^{2}\right] /\left(1-2 i^{2}\right)
=[5+6 \sqrt{2 i}+2(-1)] /(1-2(-1))\left[\right. since,\left.i^{2}=-1\right]
=[3+6 \sqrt{2 i}] / 3 =1+2 \sqrt{2 i}
As a result, the values of a, b are 1,2 \sqrt{2}

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