Question

express 1+3i/1-2i in the form of a+ib ​

Answer 1

Here Is Your Ans ⤵

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$\to\frac{1 + 3i}{1 - 2i} \\ \\ \to \frac{(1 + 3i)(1 + 2i)}{(1 - 2i)(1 + 2i)} \\ \\ \to\frac{1 + 2i + 3i + 6 {i}^{2} }{ {1}^{2} - {(2i)}^{2} } \\ \\ \to \frac{1 + 5i + 6 \times ( - 1)}{1 - 2 \times ( - 1)} \\ \\ \to \frac{1 + 5i - 6}{1 + 2} \\ \\ \to \frac{ - 5 + 5i}{3} \\ \\ \to \frac{ - 5}{3} + \frac{5i}{3} \:$

Answer 2

Answer:

$\frac{ 1+3i}{1-2i }$ in the form of with $a+ib$ can be written as $5-i$ where $a=5$ & $b=-1$

Step-by-step explanation:

Given the expression with complex numbers, $\frac{ 1+3i}{1-2i }$

Dividing and multiplying the above expression with $1+2i$,

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

Opening the brackets by multiplying the terms,

$=\frac{1\times 1+1\times3i+2i\times 1+2i\times3i}{1\times 1+1\times2i+1\times (-21)+2i\times(-2i)}$

$=\frac{1+3i+2i+6i^{2} }{1+2i -2i-4i^{2} }$

$=\frac{1+5i+6i^{2} }{1-4i^{2} }$

Using the value of $i^{2} =-1$

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

$=\frac{1+5i-6 }{1+4}$

$=\frac{5i-5 }{5}$

$=5-i$

Comparing with $a+ib$, we have $a=5$ and $b=-1$.