The modulus of the complex number $\left(\frac{3+4i}{1-2i}\right)$ is

To determine the modulus of the given complex number $\left(\frac{3+4i}{1-2i}\right)$, we must first simplify the expression into the standard form of a complex number, which is $a+bi$.

Complex Number Simplification

The given complex number is a fraction. To simplify a complex fraction, we multiply both the numerator and the denominator by the conjugate of the denominator. The denominator here is $1-2i$.

The conjugate of $1-2i$ is $1+2i$.

Let's denote the given complex number as $z$.

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

Now, we multiply the numerator and the denominator by $1+2i$:

$$z=\frac{3+4i}{1-2i}\times \frac{1+2i}{1+2i}$$

Numerator Multiplication:

We multiply the two complex numbers in the numerator: $(3+4i)(1+2i)$.

- Multiply the real parts: $3\times 1=3$
- Multiply the outer terms: $3\times 2i=6i$
- Multiply the inner terms: $4i\times 1=4i$
- Multiply the imaginary parts: $4i\times 2i=8i^2$

Combine these products:

$$(3+4i)(1+2i)=3+6i+4i+8i^2$$

Recall that $i^2=-1$. Substitute this into the expression:

$$=3+10i+8(-1)$$

$$=3+10i-8$$

$$=-5+10i$$

So, the simplified numerator is $-5+10i$.

Denominator Multiplication:

Next, we multiply the two complex numbers in the denominator: $(1-2i)(1+2i)$.

This is a product of a complex number and its conjugate, which follows the form $(a-b)(a+b)=a^2-b^2$. Here, $a=1$ and $b=2i$.

$$(1-2i)(1+2i)=(1{)}^2-(2i{)}^2$$

$$=1-4i^2$$

Again, substitute $i^2=-1$:

$$=1-4(-1)$$

$$=1+4$$

$$=5$$

So, the simplified denominator is $5$.

Combining Simplified Parts:

Now, we combine the simplified numerator and denominator to get the complex number $z$ in standard form:

$$z=\frac{-5+10i}{5}$$

Separate the real and imaginary parts:

$$z=\frac{-5}{5}+\frac{10i}{5}$$

$$z=-1+2i$$

Thus, the complex number $\left(\frac{3+4i}{1-2i}\right)$ simplifies to $-1+2i$.

Modulus Calculation

The modulus of a complex number $z=a+bi$ is calculated using the formula: $\left|z\right|=\sqrt{a^2+b^2}$.

For our simplified complex number $z=-1+2i$, we have $a=-1$ and $b=2$.

Substitute these values into the modulus formula:

$$\left|z\right|=\sqrt{{(-1)}^2+(2{)}^2}$$

Calculate the squares:

$$\left|z\right|=\sqrt{1+4}$$

Add the terms under the square root:

$$\left|z\right|=\sqrt{5}$$

Therefore, the modulus of the complex number $\left(\frac{3+4i}{1-2i}\right)$ is $\sqrt{5}$.